OP Malhotra Class 12 Maths Solutions Chapter 14 Indefinite Integral 2 Exercise 14 (A)

S Chand Class 12 ICSE Maths Solutions Chapter 14 Indefinite Integral-2 Ex 14(a)

Question 1. \( \int \frac{6 x-8}{3 x^2-8 x+5} d x \)
Answer:
Let \( I = \int \frac{6 x-8}{3 x^2-8 x+5} d x \)
We put the denominator \( 3x^2 - 8x + 5 = t \).
Now, differentiate both sides with respect to x, so \( (6x - 8) dx = dt \).
Substitute these into the integral:
\( I = \int \frac{d t}{t} \)
This is a standard integral form, which becomes \( I = \log |t| + c \).
Finally, substitute back \( t = 3x^2 - 8x + 5 \):
\( I = \log |3x^2 - 8x + 5| + c \). This method uses a direct substitution.
In simple words: We changed the bottom part of the fraction to 't' and found that the top part became 'dt'. This made the integral easy to solve as 'log t', then we just put the original expression back.

🎯 Exam Tip: When the numerator is the derivative of the denominator (or a multiple of it), a simple substitution \( t = \text{denominator} \) works perfectly, leading to a logarithmic integral.

Question 2. \( \int \frac{d x}{(3-5 x)} \)
Answer:
Let \( I = \int \frac{d x}{3-5 x} \).
We put \( 3 - 5x = t \).
Now, differentiate both sides with respect to x: \( -5 dx = dt \).
So, \( dx = \frac{dt}{-5} \).
Substitute these into the integral:
\( I = \int \frac{dt}{-5t} \)
This can be written as \( I = \frac{-1}{5} \int \frac{dt}{t} \).
Integrate the expression:
\( I = \frac{-1}{5} \log |t| + c \).
Finally, substitute back \( t = 3 - 5x \):
\( I = -\frac{1}{5} \log |3 - 5x| + c \). Remember the constant of integration for indefinite integrals.
In simple words: We used substitution for the denominator. Since the derivative of `3-5x` is `-5`, we adjusted `dx` to `dt/-5` and then solved the simple integral, putting back the original terms at the end.

🎯 Exam Tip: Always remember to divide by the coefficient of x when integrating expressions of the form \( \int \frac{1}{ax+b} dx \), which results in \( \frac{1}{a} \log |ax+b| \).

Question 3. \( \int \sqrt{1+x} d x \)
Answer:
Let \( I = \int \sqrt{1+x} d x \).
We can rewrite the square root as a power:
\( I = \int (1+x)^{\frac{1}{2}} d x \).
Now, we use the power rule for integration, \( \int u^n du = \frac{u^{n+1}}{n+1} + c \). Here, \( u = 1+x \) and \( du = dx \).
Applying the rule:
\( I = \frac{(1+x)^{\frac{1}{2}+1}}{\frac{1}{2}+1} + c \)
\( I = \frac{(1+x)^{\frac{3}{2}}}{\frac{3}{2}} + c \)
\( I = \frac{2}{3}(1+x)^{\frac{3}{2}} + c \). This is a direct application of the power rule for polynomials.
In simple words: We changed the square root into a power of one-half. Then, we used the basic rule for integrating powers: add one to the power and divide by the new power.

🎯 Exam Tip: For expressions like \( \sqrt{ax+b} \), treat them as \( (ax+b)^{1/2} \) and apply the general power rule \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + c \).

Question 4. \( \int \frac{\csc^2 x}{1+\cot x} d x \)
Answer:
Let \( I = \int \frac{\csc^2 x}{1+\cot x} d x \).
We choose to substitute the denominator: Put \( 1 + \cot x = t \).
Now, differentiate both sides with respect to x. The derivative of 1 is 0, and the derivative of \( \cot x \) is \( -\csc^2 x \).
So, \( -\csc^2 x dx = dt \).
This means \( \csc^2 x dx = -dt \).
Substitute these into the integral:
\( I = \int \frac{-dt}{t} \)
This can be written as \( I = - \int \frac{dt}{t} \).
Integrate the expression:
\( I = -\log |t| + c \).
Finally, substitute back \( t = 1 + \cot x \):
\( I = -\log |1 + \cot x| + c \). Recognizing the derivative in the numerator is key here.
In simple words: We noticed that the top part of the fraction was almost the negative derivative of the bottom part. By replacing the denominator with 't', the integral became a simple 'log t' form.

🎯 Exam Tip: Always look for a function and its derivative within the integrand. If one is the derivative of the other, a simple substitution is usually the most efficient way to solve the integral.

Question 5. \( \int \frac{(\cos x-\sin x)}{(\cos x+\sin x)} d x \)
Answer:
Let \( I = \int \frac{\cos x-\sin x}{\cos x+\sin x} d x \).
We choose to substitute the denominator: Put \( \cos x + \sin x = t \).
Now, differentiate both sides with respect to x. The derivative of \( \cos x \) is \( -\sin x \), and the derivative of \( \sin x \) is \( \cos x \).
So, \( (-\sin x + \cos x) dx = dt \), which is the same as \( (\cos x - \sin x) dx = dt \).
Substitute these into the integral:
\( I = \int \frac{dt}{t} \).
Integrate the expression:
\( I = \log |t| + c \).
Finally, substitute back \( t = \cos x + \sin x \):
\( I = \log |\cos x + \sin x| + c \). This problem is a classic example of substitution method.
In simple words: We saw that the top part of the fraction was exactly the derivative of the bottom part. So, we replaced the denominator with 't', which made the integral a simple 'log t'.

🎯 Exam Tip: This is a common pattern: \( \int \frac{f'(x)}{f(x)} dx = \log |f(x)| + C \). Always look for this form, as it simplifies integration significantly.

Question 6. \( \int \sec x \log (\sec x+\tan x) d x \)
Answer:
Let \( I = \int \sec x \log (\sec x+\tan x) d x \).
We choose to substitute the logarithmic part: Put \( \log (\sec x + \tan x) = t \).
Now, differentiate both sides with respect to x. The derivative of \( \log u \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).
So, \( \frac{1}{\sec x+\tan x} (\sec x \tan x + \sec^2 x) dx = dt \).
Factor out \( \sec x \) from the term in the parenthesis:
\( \frac{\sec x (\tan x + \sec x)}{\sec x+\tan x} dx = dt \).
The \( (\sec x + \tan x) \) terms cancel out, leaving:
\( \sec x dx = dt \).
Substitute these into the integral:
\( I = \int t \cdot dt \).
Integrate using the power rule \( \int t^n dt = \frac{t^{n+1}}{n+1} + c \):
\( I = \frac{t^2}{2} + c \).
Finally, substitute back \( t = \log (\sec x + \tan x) \):
\( I = \frac{1}{2} [\log (\sec x + \tan x)]^2 + c \). This is a helpful substitution to simplify the integral.
In simple words: We noticed that the derivative of \( \log (\sec x + \tan x) \) is \( \sec x \). So, we replaced the log part with 't' and the \( \sec x dx \) with 'dt', making the integral simple to solve.

🎯 Exam Tip: Remember the standard derivative: \( \frac{d}{dx} (\log |\sec x + \tan x|) = \sec x \). This identity is crucial for solving many integrals involving secant and tangent functions.

Question 7. \( \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x \)
Answer:
Let \( I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x \).
We choose to substitute the argument of the cosine function: Put \( \sqrt{x} = t \).
Now, differentiate both sides with respect to x. The derivative of \( \sqrt{x} \) (or \( x^{1/2} \)) is \( \frac{1}{2} x^{-1/2} \), which is \( \frac{1}{2\sqrt{x}} \).
So, \( \frac{1}{2\sqrt{x}} dx = dt \).
This means \( \frac{1}{\sqrt{x}} dx = 2 dt \).
Substitute these into the integral:
\( I = \int \cos t (2 dt) \)
\( I = 2 \int \cos t dt \).
Integrate the expression. The integral of \( \cos t \) is \( \sin t \):
\( I = 2 \sin t + c \).
Finally, substitute back \( t = \sqrt{x} \):
\( I = 2 \sin \sqrt{x} + c \). This substitution makes the integral much easier to handle.
In simple words: We replaced \( \sqrt{x} \) with 't'. We found that \( \frac{dx}{\sqrt{x}} \) turned into `2dt`. This made the integral a simple cosine function, which is easy to solve.

🎯 Exam Tip: When dealing with composite functions like \( \cos \sqrt{x} \) and the derivative of the inner function (like \( \frac{1}{\sqrt{x}} \)) is present, always try substituting the inner function to simplify.

Question 8. \( \int x e^{x^2} d x \)
Answer:
Let \( I = \int x e^{x^2} d x \).
We choose to substitute the exponent: Put \( x^2 = t \).
Now, differentiate both sides with respect to x: \( 2x dx = dt \).
This means \( x dx = \frac{dt}{2} \).
Substitute these into the integral:
\( I = \int e^t \frac{dt}{2} \)
\( I = \frac{1}{2} \int e^t dt \).
Integrate the expression. The integral of \( e^t \) is \( e^t \):
\( I = \frac{1}{2} e^t + c \).
Finally, substitute back \( t = x^2 \):
\( I = \frac{1}{2} e^{x^2} + c \). This substitution simplifies the exponent and the differential term.
In simple words: We replaced \( x^2 \) with 't' in the exponent. This made \( x dx \) become `dt/2`. The integral then became a simple 'e to the power t' form, which is very straightforward to solve.

🎯 Exam Tip: For integrals involving \( e^{f(x)} \), if \( f'(x) \) is also present in the integrand, substitute \( t = f(x) \) to simplify it to \( \int e^t dt \).

Question 9. \( \int \frac{e^{m \tan^{-1} x}}{1+x^2} d x \)
Answer:
Let \( I = \int \frac{e^{m \tan^{-1} x}}{1+x^2} d x \).
We choose to substitute the exponent: Put \( \tan^{-1} x = t \).
Now, differentiate both sides with respect to x. The derivative of \( \tan^{-1} x \) is \( \frac{1}{1+x^2} \).
So, \( \frac{1}{1+x^2} d x = dt \).
Substitute these into the integral:
\( I = \int e^{m t} d t \).
Integrate the expression. The integral of \( e^{at} \) is \( \frac{e^{at}}{a} \):
\( I = \frac{e^{m t}}{m} + c \).
Finally, substitute back \( t = \tan^{-1} x \):
\( I = \frac{e^{m \tan^{-1} x}}{m} + c \). This is a direct application of substitution to simplify the exponent.
In simple words: We replaced \( \tan^{-1} x \) with 't'. Because \( \frac{1}{1+x^2} dx \) becomes 'dt', the integral changed to a basic exponential function, which is easy to integrate.

🎯 Exam Tip: When the derivative of the inverse tangent function \( \frac{1}{1+x^2} \) appears in the integrand, it's a strong indicator to substitute \( t = \tan^{-1} x \).

Question 10. \( \int \frac{\sec^2(\log x)}{x} d x \)
Answer:
Let \( I = \int \frac{\sec^2(\log x)}{x} d x \).
We choose to substitute the argument of the secant squared function: Put \( \log x = t \).
Now, differentiate both sides with respect to x. The derivative of \( \log x \) is \( \frac{1}{x} \).
So, \( \frac{1}{x} dx = dt \).
Substitute these into the integral:
\( I = \int \sec^2 t dt \).
Integrate the expression. The integral of \( \sec^2 t \) is \( \tan t \):
\( I = \tan t + c \).
Finally, substitute back \( t = \log x \):
\( I = \tan (\log x) + c \). This substitution is very effective for simplifying the argument.
In simple words: We replaced \( \log x \) with 't'. Since \( \frac{1}{x} dx \) becomes 'dt', the integral turned into a simple \( \sec^2 t \) function, which is straightforward to solve.

🎯 Exam Tip: Always recognize that if \( \log x \) is part of a function and \( \frac{1}{x} \) is also present, substituting \( t = \log x \) is a very common and effective strategy.

Question 11. \( \int \sin^2 x \cos x d x \)
Answer:
Let \( I = \int \sin^2 x \cos x d x \).
We choose to substitute the sine function: Put \( \sin x = t \).
Now, differentiate both sides with respect to x. The derivative of \( \sin x \) is \( \cos x \).
So, \( \cos x dx = dt \).
Substitute these into the integral:
\( I = \int t^2 dt \).
Integrate using the power rule \( \int t^n dt = \frac{t^{n+1}}{n+1} + c \):
\( I = \frac{t^3}{3} + c \).
Finally, substitute back \( t = \sin x \):
\( I = \frac{\sin^3 x}{3} + c \). This is a clear case where the derivative of the base function is readily available.
In simple words: We replaced \( \sin x \) with 't'. This made \( \cos x dx \) become 'dt'. The integral then turned into a simple \( t^2 \) power function, which is easy to solve.

🎯 Exam Tip: When you see an integral of the form \( \int [f(x)]^n \cdot f'(x) dx \), the substitution \( t = f(x) \) will always simplify it to \( \int t^n dt \).

Question 12. \( \int \frac{(\sin^{-1} x)^2}{\sqrt{1-x^2}} d x \)
Answer:
Let \( I = \int \frac{(\sin^{-1} x)^2}{\sqrt{1-x^2}} d x \).
We choose to substitute the inverse sine function: Put \( \sin^{-1} x = t \).
Now, differentiate both sides with respect to x. The derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1-x^2}} \).
So, \( \frac{1}{\sqrt{1-x^2}} d x = dt \).
Substitute these into the integral:
\( I = \int t^2 dt \).
Integrate using the power rule \( \int t^n dt = \frac{t^{n+1}}{n+1} + c \):
\( I = \frac{t^3}{3} + c \).
Finally, substitute back \( t = \sin^{-1} x \):
\( I = \frac{(\sin^{-1} x)^3}{3} + c \). This substitution is very effective when dealing with inverse trigonometric functions and their derivatives.
In simple words: We replaced \( \sin^{-1} x \) with 't'. Since \( \frac{1}{\sqrt{1-x^2}} dx \) becomes 'dt', the integral became a simple \( t^2 \) power function, which is easy to solve.

🎯 Exam Tip: The derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1-x^2}} \). When this derivative is present in the integrand alongside a function of \( \sin^{-1} x \), always consider substitution with \( t = \sin^{-1} x \).

Question 13. \( \int \sec^4 x \tan x d x \)
Answer:
Let \( I = \int \sec^4 x \tan x d x \).
We can rewrite \( \sec^4 x \) as \( \sec^3 x \cdot \sec x \). So the integral becomes:
\( I = \int \sec^3 x (\sec x \tan x) d x \).
We choose to substitute \( \sec x \): Put \( \sec x = t \).
Now, differentiate both sides with respect to x. The derivative of \( \sec x \) is \( \sec x \tan x \).
So, \( \sec x \tan x d x = dt \).
Substitute these into the integral:
\( I = \int t^3 dt \).
Integrate using the power rule \( \int t^n dt = \frac{t^{n+1}}{n+1} + c \):
\( I = \frac{t^4}{4} + c \).
Finally, substitute back \( t = \sec x \):
\( I = \frac{\sec^4 x}{4} + c \). This form uses the derivative of secant effectively.
In simple words: We broke down \( \sec^4 x \) into \( \sec^3 x \) and \( \sec x \). Then, we let \( \sec x = t \), which meant \( \sec x \tan x dx \) became `dt`. The integral then became a simple \( t^3 \) power function.

🎯 Exam Tip: For powers of secant and tangent, try to isolate \( \sec^2 x \) or \( \sec x \tan x \) to use as part of \( dt \), then use identities like \( \sec^2 x = 1 + \tan^2 x \) or substitute \( \sec x \) or \( \tan x \).

Question 14. \( \int \frac{x^3}{(x^2+1)^3} d x \)
Answer:
Let \( I = \int \frac{x^3}{(x^2+1)^3} d x \).
We choose to substitute the term in the denominator: Put \( x^2 + 1 = t \).
From this, we can also say \( x^2 = t - 1 \).
Now, differentiate both sides with respect to x: \( 2x dx = dt \).
This means \( x dx = \frac{dt}{2} \).
We need to rewrite \( x^3 dx \) in terms of \( t \) and \( dt \). We can write \( x^3 dx = x^2 \cdot x dx \).
Substitute \( x^2 = t-1 \) and \( x dx = \frac{dt}{2} \):
\( I = \int \frac{(t-1)}{(t)^3} \frac{dt}{2} \)
\( I = \frac{1}{2} \int \left( \frac{t}{t^3} - \frac{1}{t^3} \right) dt \)
\( I = \frac{1}{2} \int \left( \frac{1}{t^2} - \frac{1}{t^3} \right) dt \)
\( I = \frac{1}{2} \int \left( t^{-2} - t^{-3} \right) dt \).
Integrate using the power rule \( \int t^n dt = \frac{t^{n+1}}{n+1} + c \):
\( I = \frac{1}{2} \left( \frac{t^{-2+1}}{-2+1} - \frac{t^{-3+1}}{-3+1} \right) + c \)
\( I = \frac{1}{2} \left( \frac{t^{-1}}{-1} - \frac{t^{-2}}{-2} \right) + c \)
\( I = \frac{1}{2} \left( -\frac{1}{t} + \frac{1}{2t^2} \right) + c \).
Now, substitute back \( t = x^2 + 1 \):
\( I = \frac{1}{2} \left( -\frac{1}{x^2+1} + \frac{1}{2(x^2+1)^2} \right) + c \).
We can simplify this further by finding a common denominator:
\( I = \frac{1}{2} \left( \frac{-2(x^2+1)+1}{2(x^2+1)^2} \right) + c \)
\( I = \frac{1}{2} \left( \frac{-2x^2-2+1}{2(x^2+1)^2} \right) + c \)
\( I = \frac{1}{2} \left( \frac{-2x^2-1}{2(x^2+1)^2} \right) + c \)
\( I = -\frac{2x^2+1}{4(x^2+1)^2} + c \). This problem required careful algebraic manipulation after substitution.
In simple words: We replaced \( x^2+1 \) with 't'. This helped us rewrite \( x^3 dx \) in terms of 't' and 'dt'. After splitting the fraction, we integrated each part separately using the power rule and then put the original terms back.

🎯 Exam Tip: When using substitution where the differential \( dx \) does not perfectly match, express the remaining variable terms (like \( x^2 \) in this case) in terms of the new variable \( t \) before integrating.

Question 15. \( \int \frac{1+\tan x}{x+\log \sec x} d x \)
Answer:
Let \( I = \int \frac{1+\tan x}{x+\log \sec x} d x \).
We choose to substitute the denominator: Put \( x + \log \sec x = t \).
Now, differentiate both sides with respect to x:
\( \frac{d}{dx}(x + \log \sec x) = 1 + \frac{1}{\sec x} \cdot (\sec x \tan x) \).
The \( \sec x \) terms cancel out, leaving:
\( 1 + \tan x \).
So, \( (1 + \tan x) dx = dt \).
Substitute these into the integral:
\( I = \int \frac{dt}{t} \).
Integrate the expression:
\( I = \log |t| + c \).
Finally, substitute back \( t = x + \log \sec x \):
\( I = \log |x + \log \sec x| + c \). The derivative of \( \log \sec x \) simplifies nicely to \( \tan x \).
In simple words: We let the entire bottom part be 't'. We found that the derivative of the bottom part was exactly the top part. This made the integral a simple 'log t', and then we just put the original expression back.

🎯 Exam Tip: Remember the derivative of \( \log |\sec x| = \tan x \). This is a common and useful derivative to recognize, often leading to a simple \( \int \frac{f'(x)}{f(x)} dx \) form.

Question 16. \( \int \sec^4 x d x \)
Answer:
Let \( I = \int \sec^4 x d x \).
We can rewrite \( \sec^4 x \) as \( \sec^2 x \cdot \sec^2 x \).
Using the identity \( \sec^2 x = 1 + \tan^2 x \), we get:
\( I = \int \sec^2 x (1 + \tan^2 x) d x \).
Now, we choose to substitute \( \tan x \): Put \( \tan x = t \).
Differentiate both sides with respect to x: \( \sec^2 x dx = dt \).
Substitute these into the integral:
\( I = \int (1 + t^2) dt \).
Integrate term by term:
\( I = \int 1 dt + \int t^2 dt \)
\( I = t + \frac{t^3}{3} + c \).
Finally, substitute back \( t = \tan x \):
\( I = \tan x + \frac{\tan^3 x}{3} + c \). This method smartly uses a trigonometric identity to enable substitution.
In simple words: We split \( \sec^4 x \) into \( \sec^2 x \) times \( (1 + \tan^2 x) \). Then, we replaced \( \tan x \) with 't', which made \( \sec^2 x dx \) turn into 'dt'. The integral became easy to solve as \( (1+t^2) \).

🎯 Exam Tip: When integrating even powers of \( \sec x \), always factor out \( \sec^2 x \) and convert the remaining \( \sec^2 x \) terms to \( (1+\tan^2 x) \). Then, substitute \( t = \tan x \).

Question 17. \( \int \frac{\sin (\log x)}{x} d x \)
Answer:
Let \( I = \int \frac{\sin (\log x)}{x} d x \).
We choose to substitute the argument of the sine function: Put \( \log x = t \).
Now, differentiate both sides with respect to x. The derivative of \( \log x \) is \( \frac{1}{x} \).
So, \( \frac{1}{x} dx = dt \).
Substitute these into the integral:
\( I = \int \sin t dt \).
Integrate the expression. The integral of \( \sin t \) is \( -\cos t \):
\( I = -\cos t + c \).
Finally, substitute back \( t = \log x \):
\( I = -\cos (\log x) + c \). This substitution simplifies the integral into a basic trigonometric form.
In simple words: We replaced \( \log x \) with 't'. Since \( \frac{1}{x} dx \) becomes 'dt', the integral turned into a simple \( \sin t \) function, which is easy to integrate.

🎯 Exam Tip: When \( \log x \) is inside another function (like \( \sin(\log x) \)) and \( \frac{1}{x} \) is outside, it's a clear signal to use the substitution \( t = \log x \).

Question 18. \( \int \frac{\cos x-\sin x}{1+\sin 2 x} d x \)
Answer:
Let \( I = \int \frac{\cos x-\sin x}{1+\sin 2 x} d x \).
We know the identity \( 1 + \sin 2x = 1 + 2 \sin x \cos x \).
We can also write \( 1 = \sin^2 x + \cos^2 x \).
So, \( 1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x \).
This expression is \( (\sin x + \cos x)^2 \).
Substitute this into the integral:
\( I = \int \frac{\cos x-\sin x}{(\sin x+\cos x)^2} d x \).
Now, we choose to substitute the base of the squared term: Put \( \sin x + \cos x = t \).
Differentiate both sides with respect to x: \( (\cos x - \sin x) dx = dt \).
Substitute these into the integral:
\( I = \int \frac{dt}{t^2} \).
We can rewrite this as \( I = \int t^{-2} dt \).
Integrate using the power rule \( \int t^n dt = \frac{t^{n+1}}{n+1} + c \):
\( I = \frac{t^{-2+1}}{-2+1} + c \)
\( I = \frac{t^{-1}}{-1} + c \)
\( I = -\frac{1}{t} + c \).
Finally, substitute back \( t = \sin x + \cos x \):
\( I = -\frac{1}{\sin x + \cos x} + c \). Using trigonometric identities to simplify the denominator is crucial here.
In simple words: We first changed the bottom part of the fraction, \( 1 + \sin 2x \), into \( (\sin x + \cos x)^2 \). Then, we replaced \( \sin x + \cos x \) with 't', which made the top part turn into 'dt'. This simplified the integral into \( \int t^{-2} dt \).

🎯 Exam Tip: Remember the key identity \( 1 + \sin 2x = (\sin x + \cos x)^2 \). This identity is frequently used to simplify integrands in competitive exams.

Question 19. \( \int \left(\frac{1+\sin x}{1+\cos x}\right) d x \)
Answer:
Let \( I = \int \left(\frac{1+\sin x}{1+\cos x}\right) d x \).
We can split the fraction into two parts:
\( I = \int \frac{1}{1+\cos x} d x + \int \frac{\sin x}{1+\cos x} d x \).
For the first part, use the half-angle identity \( 1+\cos x = 2 \cos^2 \frac{x}{2} \):
\( \int \frac{1}{2 \cos^2 \frac{x}{2}} d x = \int \frac{1}{2} \sec^2 \frac{x}{2} d x \).
This integrates to \( \frac{1}{2} \frac{\tan \frac{x}{2}}{\frac{1}{2}} + C_1 = \tan \frac{x}{2} + C_1 \).
For the second part, use substitution. Let \( u = 1+\cos x \). Then \( du = -\sin x dx \), so \( \sin x dx = -du \).
\( \int \frac{-du}{u} = -\log |u| + C_2 = -\log |1+\cos x| + C_2 \).
Combining both parts:
\( I = \tan \frac{x}{2} - \log |1+\cos x| + c \). Breaking down complex fractions often simplifies the problem.
In simple words: We split the fraction into two simpler integrals. For the first part, we used a half-angle identity for \( 1+\cos x \). For the second part, we used substitution by letting the denominator be 't'. Then, we combined both results.

🎯 Exam Tip: When dealing with integrals involving \( 1 \pm \sin x \) or \( 1 \pm \cos x \), consider using half-angle formulas (e.g., \( 1+\cos x = 2\cos^2(x/2) \)) or multiplying by the conjugate to simplify the expression.

Question 20. \( \int \frac{(\sin \theta+\cos \theta)}{\sqrt{\sin 2 \theta}} d \theta \)
Answer:
Let \( I = \int \frac{(\sin \theta+\cos \theta)}{\sqrt{\sin 2 \theta}} d \theta \).
This type of integral often benefits from a substitution involving \( \sin \theta - \cos \theta \) or \( \sin \theta + \cos \theta \).
Let's try: Put \( \sin \theta - \cos \theta = t \).
Now, differentiate both sides with respect to \( \theta \):
\( (\cos \theta - (-\sin \theta)) d \theta = dt \)
\( (\cos \theta + \sin \theta) d \theta = dt \). This matches the numerator.
Next, we need to express \( \sin 2 \theta \) in terms of \( t \).
Square both sides of the substitution: \( (\sin \theta - \cos \theta)^2 = t^2 \).
\( \sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta = t^2 \).
We know \( \sin^2 \theta + \cos^2 \theta = 1 \) and \( 2 \sin \theta \cos \theta = \sin 2 \theta \).
So, \( 1 - \sin 2 \theta = t^2 \).
This means \( \sin 2 \theta = 1 - t^2 \).
Substitute these into the integral:
\( I = \int \frac{dt}{\sqrt{1 - t^2}} \).
This is a standard integral form, which is \( \sin^{-1} t \):
\( I = \sin^{-1} t + c \).
Finally, substitute back \( t = \sin \theta - \cos \theta \):
\( I = \sin^{-1} (\sin \theta - \cos \theta) + c \). This is a common trick for integrals involving \( \sin 2 \theta \).
In simple words: We made a clever substitution, letting \( \sin \theta - \cos \theta \) be 't'. This made the top part turn into 'dt' and the \( \sin 2 \theta \) under the root became \( 1 - t^2 \). The integral then became a simple \( \sin^{-1} t \) form.

🎯 Exam Tip: For integrals containing \( \sqrt{\sin 2x} \) or similar forms, consider substitutions like \( t = \sin x \pm \cos x \). Squaring \( t \) will often help convert \( \sin 2x \) into terms of \( t \).

Question 21. \( \int \frac{\cos x}{(\cos \frac{x}{2}+\sin \frac{x}{2})^2} d x \)
Answer:
Let \( I = \int \frac{\cos x}{(\cos \frac{x}{2}+\sin \frac{x}{2})^2} d x \).
First, simplify the denominator using the identity \( (\cos A + \sin A)^2 = \cos^2 A + \sin^2 A + 2 \sin A \cos A = 1 + \sin 2A \).
So, \( (\cos \frac{x}{2}+\sin \frac{x}{2})^2 = 1 + \sin (2 \cdot \frac{x}{2}) = 1 + \sin x \).
The integral becomes \( I = \int \frac{\cos x}{1+\sin x} d x \).
Now, we choose to substitute the denominator: Put \( 1 + \sin x = t \).
Differentiate both sides with respect to x: \( \cos x dx = dt \).
Substitute these into the integral:
\( I = \int \frac{dt}{t} \).
Integrate the expression:
\( I = \log |t| + c \).
Finally, substitute back \( t = 1 + \sin x \):
\( I = \log |1 + \sin x| + c \). This problem showcases the effective use of a trigonometric identity.
In simple words: We first simplified the bottom part of the fraction using a special trigonometric identity, turning it into \( 1 + \sin x \). Then, we let \( 1 + \sin x \) be 't', and because \( \cos x dx \) became 'dt', the integral became a simple 'log t' form.

🎯 Exam Tip: Recognizing the expansion of \( (\cos(x/2) + \sin(x/2))^2 \) as \( 1+\sin x \) is a common simplification technique in integration problems involving half-angles.

Question 22. \( \int \frac{\sin x \cos x d x}{\sin^4 x+\cos^4 x} \)
Answer:
Let \( I = \int \frac{\sin x \cos x d x}{\sin^4 x+\cos^4 x} \).
To simplify this integral, divide both the numerator and denominator by \( \cos^4 x \):
\( I = \int \frac{\frac{\sin x \cos x}{\cos^4 x}}{\frac{\sin^4 x}{\cos^4 x}+\frac{\cos^4 x}{\cos^4 x}} d x \)
\( I = \int \frac{\frac{\sin x}{\cos x} \cdot \frac{1}{\cos^2 x}}{\tan^4 x+1} d x \)
\( I = \int \frac{\tan x \sec^2 x}{\tan^4 x+1} d x \).
Now, we choose to substitute \( \tan^2 x \): Put \( \tan^2 x = t \).
Differentiate both sides with respect to x: \( 2 \tan x \sec^2 x dx = dt \).
This means \( \tan x \sec^2 x dx = \frac{dt}{2} \).
Substitute these into the integral:
\( I = \int \frac{\frac{dt}{2}}{t^2+1} \)
\( I = \frac{1}{2} \int \frac{dt}{t^2+1} \).
This is a standard integral form, which is \( \tan^{-1} t \):
\( I = \frac{1}{2} \tan^{-1} t + c \).
Finally, substitute back \( t = \tan^2 x \):
\( I = \frac{1}{2} \tan^{-1} (\tan^2 x) + c \). Dividing by the highest power of cosine is a standard method here.
In simple words: We divided the top and bottom of the fraction by \( \cos^4 x \) to get terms with \( \tan x \) and \( \sec^2 x \). Then, we let \( \tan^2 x \) be 't', which made the top part turn into \( \frac{1}{2} dt \). The integral became a simple \( \tan^{-1} t \) form.

🎯 Exam Tip: For integrals involving combinations of \( \sin x \) and \( \cos x \) in powers, especially when the sum of powers in the numerator is less than or equal to that in the denominator, dividing by a suitable power of \( \cos x \) (or \( \sin x \)) to convert to \( \tan x \) and \( \sec x \) is often effective.

Question 23. \( \int \frac{\tan x}{\sec x+\tan x} d x \)
Answer:
Let \( I = \int \frac{\tan x}{\sec x+\tan x} d x \).
Multiply the numerator and denominator by the conjugate of the denominator, \( (\sec x - \tan x) \):
\( I = \int \frac{\tan x (\sec x - \tan x)}{(\sec x+\tan x)(\sec x - \tan x)} d x \)
In the denominator, \( (\sec x+\tan x)(\sec x - \tan x) = \sec^2 x - \tan^2 x \).
Using the identity \( \sec^2 x - \tan^2 x = 1 \).
So, the integral simplifies to:
\( I = \int \tan x (\sec x - \tan x) d x \)
\( I = \int (\sec x \tan x - \tan^2 x) d x \).
Now, we integrate term by term. We know \( \int \sec x \tan x d x = \sec x \).
For the second term, use the identity \( \tan^2 x = \sec^2 x - 1 \):
\( I = \int \sec x \tan x d x - \int (\sec^2 x - 1) d x \)
\( I = \int \sec x \tan x d x - \int \sec^2 x d x + \int 1 d x \).
Integrate each term:
\( I = \sec x - \tan x + x + c \). Multiplying by the conjugate is a common technique for such integrals.
In simple words: We multiplied the top and bottom of the fraction by \( \sec x - \tan x \). This made the bottom part equal to 1. Then we broke the integral into two parts. One part was \( \sec x \tan x \), which integrates to \( \sec x \). The other part was \( \tan^2 x \), which we changed to \( \sec^2 x - 1 \) before integrating.

🎯 Exam Tip: When the integrand involves \( \sec x + \tan x \) in the denominator, multiplying by its conjugate \( \sec x - \tan x \) is a standard technique. This simplifies the denominator to 1, making the integration much simpler.

Question 24. \( \int \tan^4 x d x \)
Answer:
Let \( I = \int \tan^4 x d x \).
We can rewrite \( \tan^4 x \) as \( \tan^2 x \cdot \tan^2 x \).
Using the identity \( \tan^2 x = \sec^2 x - 1 \), substitute one of the \( \tan^2 x \) terms:
\( I = \int \tan^2 x (\sec^2 x - 1) d x \)
\( I = \int (\tan^2 x \sec^2 x - \tan^2 x) d x \).
Now, integrate term by term:
\( I = \int \tan^2 x \sec^2 x d x - \int \tan^2 x d x \).
For the first integral, we can use substitution. Let \( u = \tan x \). Then \( du = \sec^2 x dx \).
\( \int u^2 du = \frac{u^3}{3} = \frac{\tan^3 x}{3} \).
For the second integral, again use the identity \( \tan^2 x = \sec^2 x - 1 \):
\( \int \tan^2 x d x = \int (\sec^2 x - 1) d x = \int \sec^2 x d x - \int 1 d x = \tan x - x \).
Combining both results:
\( I = \frac{\tan^3 x}{3} - (\tan x - x) + c \)
\( I = \frac{\tan^3 x}{3} - \tan x + x + c \). Repeated use of trigonometric identities simplifies this.
In simple words: We broke \( \tan^4 x \) into \( \tan^2 x \) times \( (\sec^2 x - 1) \). Then we separated it into two integrals. The first integral used substitution for \( \tan x \), and the second integral needed another rewrite of \( \tan^2 x \) to be solved.

🎯 Exam Tip: When integrating even powers of \( \tan x \), peel off \( \tan^2 x \) and replace it with \( \sec^2 x - 1 \). This creates terms that are directly integrable or solvable by simple substitution.

Question 25. \( \int \frac{d x}{\sqrt{1+\sqrt{x}}} \)
Answer:
Let \( I = \int \frac{d x}{\sqrt{1+\sqrt{x}}} \).
We choose to substitute the entire nested square root: Put \( \sqrt{1+\sqrt{x}} = t \).
Square both sides: \( 1+\sqrt{x} = t^2 \).
Then, \( \sqrt{x} = t^2 - 1 \).
Square both sides again: \( x = (t^2 - 1)^2 \).
Now, differentiate \( x \) with respect to \( t \) to find \( dx \):
\( \frac{dx}{dt} = 2(t^2 - 1) \cdot (2t) \)
\( dx = 4t(t^2 - 1) dt \).
Substitute these into the integral:
\( I = \int \frac{4t(t^2 - 1) dt}{t} \)
The 't' in the numerator and denominator cancels:
\( I = \int 4(t^2 - 1) dt \)
\( I = 4 \int (t^2 - 1) dt \).
Integrate term by term:
\( I = 4 \left( \frac{t^3}{3} - t \right) + c \).
Finally, substitute back \( t = \sqrt{1+\sqrt{x}} \):
\( I = 4 \left( \frac{(\sqrt{1+\sqrt{x}})^3}{3} - \sqrt{1+\sqrt{x}} \right) + c \). This complex substitution is very effective here.
In simple words: We replaced \( \sqrt{1+\sqrt{x}} \) with 't'. This required us to square both sides twice to express 'x' in terms of 't'. Then we found 'dx' and substituted everything into the integral, which simplified it to a basic polynomial to integrate.

🎯 Exam Tip: For integrals involving nested square roots, substitute the outermost or most complex root. This may require multiple steps of squaring and differentiation to express \( dx \) correctly.

Question 26. \( \int \frac{x d x}{\sqrt{x+5}} \)
Answer:
Let \( I = \int \frac{x d x}{\sqrt{x+5}} \).
We choose to substitute the term under the square root: Put \( x+5 = t \).
From this, we can also say \( x = t-5 \).
Differentiate both sides with respect to x: \( dx = dt \).
Substitute these into the integral:
\( I = \int \frac{(t-5) dt}{\sqrt{t}} \)
\( I = \int \left( \frac{t}{\sqrt{t}} - \frac{5}{\sqrt{t}} \right) dt \)
\( I = \int (t^{1/2} - 5t^{-1/2}) dt \).
Integrate term by term using the power rule \( \int t^n dt = \frac{t^{n+1}}{n+1} + c \):
\( I = \frac{t^{1/2+1}}{1/2+1} - 5 \frac{t^{-1/2+1}}{-1/2+1} + c \)
\( I = \frac{t^{3/2}}{3/2} - 5 \frac{t^{1/2}}{1/2} + c \)
\( I = \frac{2}{3} t^{3/2} - 10 t^{1/2} + c \).
Finally, substitute back \( t = x+5 \):
\( I = \frac{2}{3} (x+5)^{3/2} - 10 (x+5)^{1/2} + c \). This substitution is helpful when a term in the numerator is closely related to the term in the square root.
In simple words: We replaced \( x+5 \) with 't', so 'x' became \( t-5 \). This allowed us to rewrite the integral in terms of 't' and simplify it into powers of 't' that are easy to integrate.

🎯 Exam Tip: For integrals like \( \int \frac{x}{ \sqrt{ax+b}} dx \) or \( \int x \sqrt{ax+b} dx \), substitute \( t = ax+b \). This helps convert both \( x \) and \( dx \) into terms of \( t \), simplifying the integrand significantly.

Question 27. \( \int \frac{(x+1)(x+\log x)^2}{x} d x \)
Answer:
Let \( I = \int \frac{(x+1)(x+\log x)^2}{x} d x \).
We can rewrite the expression as \( I = \int \left(1+\frac{1}{x}\right) (x+\log x)^2 d x \).
Now, we choose to substitute the base of the squared term: Put \( x+\log x = t \).
Differentiate both sides with respect to x: \( \left(1+\frac{1}{x}\right) dx = dt \).
Substitute these into the integral:
\( I = \int t^2 dt \).
Integrate using the power rule \( \int t^n dt = \frac{t^{n+1}}{n+1} + c \):
\( I = \frac{t^3}{3} + c \).
Finally, substitute back \( t = x+\log x \):
\( I = \frac{(x+\log x)^3}{3} + c \). Recognizing the derivative of \( x+\log x \) is the key here.
In simple words: We grouped the terms to see that \( \left(1+\frac{1}{x}\right) \) is the derivative of \( (x+\log x) \). So, we replaced \( x+\log x \) with 't' and \( \left(1+\frac{1}{x}\right) dx \) with 'dt', making the integral a simple \( t^2 \) power function.

🎯 Exam Tip: For integrals of the form \( \int [f(x)]^n \cdot f'(x) dx \), a substitution of \( t = f(x) \) simplifies the integral to \( \int t^n dt \). Always be on the lookout for a function and its derivative.

Question 28. \( \int x^2 e^{x^3} \cos \left(e^{x^3}\right) d x \)
Answer:
Let \( I = \int x^2 e^{x^3} \cos \left(e^{x^3}\right) d x \).
We choose to substitute the argument of the cosine function, which is also the exponent of 'e': Put \( e^{x^3} = t \).
Differentiate both sides with respect to x using the chain rule. The derivative of \( e^u \) is \( e^u \frac{du}{dx} \).
So, \( e^{x^3} \cdot (3x^2) dx = dt \).
This means \( x^2 e^{x^3} dx = \frac{dt}{3} \).
Substitute these into the integral:
\( I = \int \cos t \frac{dt}{3} \)
\( I = \frac{1}{3} \int \cos t dt \).
Integrate the expression. The integral of \( \cos t \) is \( \sin t \):
\( I = \frac{1}{3} \sin t + c \).
Finally, substitute back \( t = e^{x^3} \):
\( I = \frac{1}{3} \sin \left(e^{x^3}\right) + c \). This chained substitution effectively simplifies the complex expression.
In simple words: We replaced \( e^{x^3} \) with 't'. This made \( x^2 e^{x^3} dx \) turn into \( \frac{1}{3} dt \). The integral then became a simple \( \cos t \) function, which is easy to solve.

🎯 Exam Tip: When an integral has a complex function like \( e^{x^3} \) nested inside another function (like \( \cos \)), and its derivative (or part of it) is also present, a layered substitution (like \( t=e^{x^3} \)) is often the correct approach.

Question 29. \( \int \frac{2 \sin \theta \cos \theta}{\sin^4 \theta+\cos^4 \theta} d \theta \)
Answer:
Let \( I = \int \frac{2 \sin \theta \cos \theta}{\sin^4 \theta+\cos^4 \theta} d \theta \).
First, simplify the numerator: \( 2 \sin \theta \cos \theta = \sin 2 \theta \).
Divide both the numerator and denominator by \( \cos^4 \theta \):
\( I = \int \frac{\frac{2 \sin \theta \cos \theta}{\cos^4 \theta}}{\frac{\sin^4 \theta}{\cos^4 \theta}+\frac{\cos^4 \theta}{\cos^4 \theta}} d \theta \)
\( I = \int \frac{2 \tan \theta \sec^2 \theta}{\tan^4 \theta+1} d \theta \).
Now, we choose to substitute \( \tan^2 \theta \): Put \( \tan^2 \theta = t \).
Differentiate both sides with respect to \( \theta \): \( 2 \tan \theta \sec^2 \theta d \theta = dt \).
Substitute these into the integral:
\( I = \int \frac{dt}{t^2+1} \).
This is a standard integral form, which is \( \tan^{-1} t \):
\( I = \tan^{-1} t + c \).
Finally, substitute back \( t = \tan^2 \theta \):
\( I = \tan^{-1} (\tan^2 \theta) + c \). This problem is similar to Q22, using the same strategy.
In simple words: We divided the top and bottom of the fraction by \( \cos^4 \theta \) to get terms with \( \tan \theta \) and \( \sec^2 \theta \). Then, we let \( \tan^2 \theta \) be 't', which made the numerator turn into 'dt'. The integral became a simple \( \tan^{-1} t \) form.

🎯 Exam Tip: For integrals with similar powers of sine and cosine in the denominator, try converting the expression into terms of \( \tan \) and \( \sec \) by dividing by the highest power of \( \cos \).

Question 30. \( \int \frac{1}{9+16 \cos^2 x} d x \)
Answer:
Let \( I = \int \frac{1}{9+16 \cos^2 x} d x \).
To simplify, divide both the numerator and denominator by \( \cos^2 x \):
\( I = \int \frac{\frac{1}{\cos^2 x}}{\frac{9}{\cos^2 x}+\frac{16 \cos^2 x}{\cos^2 x}} d x \)
\( I = \int \frac{\sec^2 x}{9 \sec^2 x+16} d x \).
Now, use the identity \( \sec^2 x = 1 + \tan^2 x \) in the denominator:
\( I = \int \frac{\sec^2 x}{9(1+\tan^2 x)+16} d x \)
\( I = \int \frac{\sec^2 x}{9+9 \tan^2 x+16} d x \)
\( I = \int \frac{\sec^2 x}{25+9 \tan^2 x} d x \).
Now, we choose to substitute \( \tan x \): Put \( \tan x = t \).
Differentiate both sides with respect to x: \( \sec^2 x dx = dt \).
Substitute these into the integral:
\( I = \int \frac{dt}{25+9 t^2} \).
We can factor out 9 from the denominator:
\( I = \int \frac{dt}{9(\frac{25}{9}+t^2)} = \frac{1}{9} \int \frac{dt}{t^2+(\frac{5}{3})^2} \).
This is a standard integral form \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + c \). Here, \( a = \frac{5}{3} \).
\( I = \frac{1}{9} \cdot \frac{1}{\frac{5}{3}} \tan^{-1} \left(\frac{t}{\frac{5}{3}}\right) + c \)
\( I = \frac{1}{9} \cdot \frac{3}{5} \tan^{-1} \left(\frac{3t}{5}\right) + c \)
\( I = \frac{1}{15} \tan^{-1} \left(\frac{3t}{5}\right) + c \).
Finally, substitute back \( t = \tan x \):
\( I = \frac{1}{15} \tan^{-1} \left(\frac{3 \tan x}{5}\right) + c \). This type of transformation to \( \tan x \) is very useful.
In simple words: We divided the top and bottom by \( \cos^2 x \) to get \( \sec^2 x \) and \( \tan^2 x \) terms. Then, we let \( \tan x \) be 't'. This transformed the integral into a standard \( \frac{1}{a^2+x^2} \) form, which we solved using the inverse tangent formula.

🎯 Exam Tip: For integrals of the form \( \int \frac{1}{a+b \cos^2 x} dx \) or \( \int \frac{1}{a+b \sin^2 x} dx \), divide the numerator and denominator by \( \cos^2 x \) to convert the expression into terms of \( \tan x \) and \( \sec^2 x \), then use the substitution \( t = \tan x \).

Question 31. \( \int \frac{\sin x}{a+b \cos x} d x \)
Answer: To solve this integral, we use the substitution method. Let \( t = a + b \cos x \). When we differentiate \( t \) with respect to \( x \), we get \( dt = -b \sin x \, dx \). From this, we can write \( \sin x \, dx = -\frac{1}{b} dt \). Substituting these into the integral, it becomes \( \int \frac{-\frac{1}{b} dt}{t} = -\frac{1}{b} \int \frac{1}{t} dt \). Integrating \( \frac{1}{t} \) gives \( \log|t| \). So, the solution is \( -\frac{1}{b} \log|t| + c \). Finally, we substitute \( t \) back with \( a + b \cos x \) to get the final answer: \( -\frac{1}{b} \log|a + b \cos x| + c \). This integral is a classic example of substitution method where the derivative of the denominator is present in the numerator (with a constant multiple).
In simple words: We let the bottom part of the fraction, \( a + b \cos x \), be \( t \). When we find `dt`, we see `sin x dx` in it, so we replace `sin x dx` with `(-1/b) dt`. The integral then turns into `(-1/b) \int (1/t) dt`, which is `(-1/b) log|t|`. Finally, we put `a + b cos x` back for `t`.

🎯 Exam Tip: Recognize integrals of the form \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C \). This method often involves substitution, where the derivative of the denominator (or a part of it) is present in the numerator.

Question 32. \( \int \frac{e^x-1}{e^x+1} d x \)
Answer: To evaluate this integral, we first manipulate the numerator by adding and subtracting 1: \( e^x-1 = (e^x+1)-2 \). This allows us to split the fraction into two simpler integrals: \( \int \frac{e^x+1}{e^x+1} dx - \int \frac{2}{e^x+1} dx \). The first integral simplifies to \( \int 1 dx = x \). For the second integral, \( 2 \int \frac{1}{e^x+1} dx \), we multiply the numerator and denominator by \( e^{-x} \) to make a substitution easier: \( 2 \int \frac{e^{-x}}{e^{-x}(e^x+1)} dx = 2 \int \frac{e^{-x}}{1+e^{-x}} dx \). Now, let \( t = 1+e^{-x} \). Differentiating \( t \) gives \( dt = -e^{-x} dx \), so \( e^{-x} dx = -dt \). Substituting this into the second integral, we get \( 2 \int \frac{-dt}{t} = -2 \log|t| \). Combining everything, the solution is \( x - (-2 \log|1+e^{-x}|) + c = x + 2 \log|1+e^{-x}| + c \). Manipulating the expression is key to solving this type of integral.
In simple words: First, we change the top part of the fraction to `(e^x+1) - 2` and split the integral into two parts. The first part becomes `x`. For the second part, we multiply the top and bottom by `e^{-x}`. Then, we let `t` be `1+e^{-x}`. This helps us integrate to `-2 log|t|`. Finally, we put `1+e^{-x}` back for `t` and combine both parts.

🎯 Exam Tip: For integrals like \( \int \frac{e^x-1}{e^x+1} dx \), a common trick is to rewrite the numerator as \( (e^x+1)-2 \) or divide both numerator and denominator by \( e^x \). This helps simplify the expression for easier integration.

Question 33. \( \int \frac{e^{2x}}{e^{2 x}-2} d x \)
Answer: To solve this integral, we use a simple substitution method. We set \( t \) equal to the denominator, \( e^{2x}-2 \). Next, we differentiate \( t \) with respect to \( x \): \( dt = \frac{d}{dx}(e^{2x}-2) dx = (2e^{2x}) dx \). From this, we can express \( e^{2x} dx \) as \( \frac{1}{2} dt \). Substituting these into the original integral, it transforms into \( \int \frac{\frac{1}{2} dt}{t} = \frac{1}{2} \int \frac{1}{t} dt \). This is a standard integral, which evaluates to \( \frac{1}{2} \log|t| \). Finally, we replace \( t \) with its original expression, \( e^{2x}-2 \), to get the complete solution: \( \frac{1}{2} \log|e^{2x}-2| + c \). This method is very efficient when the numerator is a constant multiple of the derivative of the denominator.
In simple words: We let the bottom part, `e^{2x}-2`, be `t`. When we find `dt`, we see `e^{2x} dx` in it. So we replace `e^{2x} dx` with `dt/2`. Then we integrate `1/(2t)`, which gives `(1/2)log|t|`. In the end, we put `e^{2x}-2` back instead of `t`.

🎯 Exam Tip: Look for integrals where the numerator is a multiple of the derivative of the denominator. In such cases, a simple substitution with `t = \text{denominator}` often simplifies the integral to `\int \frac{1}{t} dt` or a similar form.

Question 34. \( \int \frac{e^{2 x}}{e^x-1} d x \)
Answer: To evaluate this integral, we use a substitution method. We set \( t = e^x-1 \). From this, we can also write \( e^x = t+1 \). Next, we find the differential \( dt \) by differentiating \( t \) with respect to \( x \): \( dt = e^x dx \). We can rewrite the numerator \( e^{2x} \) as \( e^x \cdot e^x \). Substituting all these into the integral, it becomes \( \int \frac{(t+1) dt}{t} \), since \( e^x \) is \( t+1 \) and \( e^x dx \) is \( dt \). This integral can be split into two simpler terms: \( \int (1 + \frac{1}{t}) dt \). Integrating term by term, we get \( t + \log|t| + c \). Finally, we substitute \( t \) back with \( e^x-1 \) to express the answer in terms of \( x \): \( e^x + \log|e^x-1| + c \). This method is useful for integrals involving exponential functions where a simple substitution simplifies the expression.
In simple words: We let `e^x-1` be `t`. This helps us replace `e^x` with `t+1` and `e^x dx` with `dt`. We also change `e^{2x}` to `e^x \cdot e^x`. After putting `t` into the integral, we get a simpler sum of `1` and `1/t`. We integrate these parts, which gives `t` and `log|t|`. Then, we change `t` back to `e^x-1`.

🎯 Exam Tip: When dealing with \( e^x \) in integrals, substituting \( t = e^x \) or \( t = e^x \pm \text{constant} \) is often effective. Always remember to also express \( dx \) in terms of \( dt \) and \( t \).

Question 35. \( \int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x \)
Answer: To solve this integral, we use a direct substitution method. We let \( t \) be the entire denominator: \( t = e^{2x} + e^{-2x} \). Next, we find the differential \( dt \) by differentiating \( t \) with respect to \( x \). The derivative of \( e^{2x} \) is \( 2e^{2x} \), and the derivative of \( e^{-2x} \) is \( -2e^{-2x} \). So, \( dt = (2e^{2x} - 2e^{-2x}) dx \). We can factor out a 2: \( dt = 2(e^{2x} - e^{-2x}) dx \). This means the numerator, \( (e^{2x} - e^{-2x}) dx \), can be replaced by \( \frac{1}{2} dt \). Substituting these into the integral, it becomes \( \int \frac{\frac{1}{2} dt}{t} = \frac{1}{2} \int \frac{1}{t} dt \). Integrating \( \frac{1}{t} \) gives \( \log|t| \). So, the solution is \( \frac{1}{2} \log|t| + c \). Finally, we substitute \( t \) back with \( e^{2x} + e^{-2x} \) to obtain the answer: \( \frac{1}{2} \log|e^{2x} + e^{-2x}| + c \). This is a straightforward example of identifying the derivative of the denominator in the numerator.
In simple words: We can solve this integral by letting `t` be the whole bottom part, `e^{2x} + e^{-2x}`. When we find the derivative of `t`, we notice it looks very similar to the top part of the fraction, just needing a division by `2`. So, the integral becomes `(1/2) \int (1/t) dt`, which is easy to solve as `(1/2)log|t|`. Then, we put `e^{2x} + e^{-2x}` back for `t`.

🎯 Exam Tip: This integral is a direct application of the \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C \) rule. Always check if the numerator is the derivative (or a constant multiple of the derivative) of the denominator.

Question 36. \( \int \cot^3 x \operatorname{cosec}^4 x d x \)
Answer: To evaluate this integral, we first rewrite \( \operatorname{cosec}^4 x \) as \( \operatorname{cosec}^2 x \cdot \operatorname{cosec}^2 x \). Using the trigonometric identity \( \operatorname{cosec}^2 x = 1 + \cot^2 x \), we substitute one \( \operatorname{cosec}^2 x \) term, making the expression \( \int \cot^3 x (1+\cot^2 x) \operatorname{cosec}^2 x dx \). This setup is ideal for a substitution. We let \( t = \cot x \). Differentiating \( t \) gives \( dt = -\operatorname{cosec}^2 x dx \), so \( \operatorname{cosec}^2 x dx = -dt \). Substituting \( t \) and \( -dt \) into the integral transforms it into \( \int t^3 (1+t^2) (-dt) = -\int (t^3+t^5) dt \). Now, we integrate term by term using the power rule for integration: \( -(\frac{t^4}{4} + \frac{t^6}{6}) + c \). Finally, we substitute \( t \) back with \( \cot x \) to obtain the result in terms of \( x \): \( -\frac{\cot^4 x}{4} - \frac{\cot^6 x}{6} + c \). This strategy helps convert complex trigonometric integrals into simpler polynomial forms.
In simple words: First, we change `cosec⁴ x` to `cosec² x` times `(1+cot² x)`. Then, we let `cot x` be `t`. This makes `cosec² x dx` turn into `-dt`. After substituting, we get a simpler integral with `t³(1+t²)`. We multiply this out to `t³+t⁵` and integrate each part. At last, we put `cot x` back in place of `t`.

🎯 Exam Tip: When integrating powers of `cot x` and `cosec x`, or `tan x` and `sec x`, try to separate one `cosec² x` (or `sec² x`) and convert the remaining `cosec x` (or `sec x`) terms into `cot x` (or `tan x`) using identities. Then, use `t = cot x` (or `t = tan x`).

Question 37. \( \int \sqrt{e^x-4} d x \)
Answer: To solve this integral, we use a substitution designed to eliminate the square root. We let \( t = \sqrt{e^x-4} \). Squaring both sides gives \( t^2 = e^x-4 \), which implies \( e^x = t^2+4 \). Now, we differentiate \( e^x = t^2+4 \) with respect to \( x \): \( e^x dx = 2t dt \). From this, we can find \( dx = \frac{2t dt}{e^x} \). Substituting \( e^x = t^2+4 \), we get \( dx = \frac{2t dt}{t^2+4} \). Now, we substitute \( t \) and \( dx \) back into the original integral: \( \int t \cdot \frac{2t dt}{t^2+4} = \int \frac{2t^2 dt}{t^2+4} \). To integrate this, we perform algebraic manipulation on the numerator: \( \int \frac{2(t^2+4-4)}{t^2+4} dt = \int (2 - \frac{8}{t^2+4}) dt \). We split this into two integrals: \( \int 2 dt - 8 \int \frac{1}{t^2+2^2} dt \). Integrating gives \( 2t - 8 \cdot \frac{1}{2} \tan^{-1}(\frac{t}{2}) + c = 2t - 4 \tan^{-1}(\frac{t}{2}) + c \). Finally, we substitute \( t \) back with \( \sqrt{e^x-4} \) to get the solution: \( 2\sqrt{e^x-4} - 4 \tan^{-1}(\frac{\sqrt{e^x-4}}{2}) + c \). This method effectively converts a complex integral into a sum of simpler standard forms.
In simple words: We start by making `t` equal to the square root part, `\sqrt{e^x-4}`. Then, we find `dx` in terms of `t`. This changes the whole integral to be about `t`. The new integral looks like `\int \frac{2t²}{t²+4} dt`. We cleverly rewrite `2t²` as `2(t²+4-4)` to simplify the fraction. This gives us `2 - \frac{8}{t²+4}`. We then integrate `2` and the `tan^{-1}` form. At the end, we put `\sqrt{e^x-4}` back in place of `t`.

🎯 Exam Tip: When an integral contains a square root like \( \sqrt{e^x-k} \), a common strategy is to substitute \( t = \sqrt{\text{expression}} \). This helps remove the square root and convert the integral into a rational function of \( t \), which can then be solved using standard techniques.

Question 38. \( \int \frac{x d x}{e^{x^2}} \)
Answer: To solve this integral, we first rewrite the expression by moving the exponential term from the denominator to the numerator: \( \int x e^{-x^2} dx \). Now, we use the method of substitution. We let \( t = x^2 \). Differentiating \( t \) with respect to \( x \) gives \( dt = 2x dx \). From this, we can write \( x dx = \frac{1}{2} dt \). Substituting these into the integral, it transforms into \( \int e^{-t} \frac{1}{2} dt = \frac{1}{2} \int e^{-t} dt \). The integral of \( e^{-t} \) is \( -e^{-t} \). So, the solution is \( \frac{1}{2} (-e^{-t}) + c = -\frac{1}{2} e^{-t} + c \). Finally, we substitute \( t \) back with \( x^2 \) to get the answer in terms of \( x \): \( -\frac{1}{2} e^{-x^2} + c \). This is a good example of how recognizing a function and its derivative within the integrand simplifies the problem greatly.
In simple words: First, we change `1/e^{x^2}` to `e^{-x^2}`. Then, we let `x²` be `t`. When we find `dt`, we see `x dx` in it, which helps us replace `x dx` with `dt/2`. The integral then becomes `(1/2) \int e^{-t} dt`. We know how to integrate `e^{-t}`, which gives `-e^{-t}`. Finally, we put `x²` back for `t`.

🎯 Exam Tip: For integrals involving \( e^{f(x)} \), try substituting \( t = f(x) \) if \( f'(x) \) or a multiple of \( f'(x) \) is also present in the integrand. This often simplifies the integral to \( \int e^t dt \).

Question 40. \( \int\left(x^3-1\right)^{1 / 3} x^5 d x \)
Answer: To solve this integral, we use a substitution method. We set \( t = (x^3-1)^{1/3} \). To remove the fractional exponent, we cube both sides, which gives \( t^3 = x^3-1 \). From this, we can express \( x^3 = t^3+1 \). Next, we differentiate \( t^3 = x^3-1 \) with respect to \( x \): \( 3t^2 dt = 3x^2 dx \), which simplifies to \( x^2 dx = t^2 dt \). To prepare the original integral for substitution, we split \( x^5 \) into \( x^3 \cdot x^2 \). Now, substituting all parts in terms of \( t \): \( \int (x^3-1)^{1/3} \cdot x^3 \cdot x^2 dx = \int t \cdot (t^3+1) \cdot t^2 dt \). This simplifies to \( \int t^3(t^3+1) dt = \int (t^6+t^3) dt \). This is a simple polynomial integral. Integrating term by term using the power rule, we get \( \frac{t^7}{7} + \frac{t^4}{4} + c \). Finally, we substitute \( t \) back with \( (x^3-1)^{1/3} \) to express the answer in terms of \( x \): \( \frac{(x^3-1)^{7/3}}{7} + \frac{(x^3-1)^{4/3}}{4} + c \). This technique of breaking down polynomial terms into factors for substitution is quite effective.
In simple words: We start by making `t` equal to `(x³-1)^{1/3}`. Then, we find `x³` in terms of `t` and `x² dx` in terms of `t dt`. We split `x⁵` into `x³ \cdot x²`. After putting all these `t` parts into the integral, it becomes `\int t³(t³+1) dt`. We multiply this out to `t⁶+t³` and integrate each term. At the very end, we put `(x³-1)^{1/3}` back for `t`.

🎯 Exam Tip: When integrating expressions involving \( (ax^n+b)^m \), it's often helpful to substitute \( t = (ax^n+b) \). If \( x \) outside the parenthesis has a power that is \( n-1 \) or \( 2n-1 \) (or related), this substitution will often work. Break \( x^k \) into suitable powers like \( x^n \cdot x^{n-1} \) if needed.

Question 41. \( \int \sin ^3 x \cos ^4 x d x \)
Answer: To solve this trigonometric integral, we first rewrite \( \sin^3 x \) as \( \sin x \cdot \sin^2 x \). Then, using the Pythagorean identity \( \sin^2 x = 1 - \cos^2 x \), we transform the expression into \( \int \sin x (1-\cos^2 x) \cos^4 x dx \). This setup is perfect for a substitution. We let \( t = \cos x \). Differentiating \( t \) with respect to \( x \) gives \( dt = -\sin x dx \), so \( \sin x dx = -dt \). Substituting \( t \) and \( -dt \) into the integral, it becomes \( -\int (1-t^2) t^4 dt \). Now, we expand the integrand: \( -\int (t^4 - t^6) dt \). We integrate each term using the power rule: \( -(\frac{t^5}{5} - \frac{t^7}{7}) + c \). Finally, we substitute \( t \) back with \( \cos x \) to get the solution in terms of \( x \): \( -\frac{\cos^5 x}{5} + \frac{\cos^7 x}{7} + c \). This technique is a standard approach for integrals involving powers of sine and cosine where one function has an odd power.
In simple words: We start by changing `\sin³ x` to `\sin x` times `(1-\cos² x)`. Then, we let `\cos x` be `t`. This makes `\sin x dx` turn into `-dt`. After putting `t` into the integral, it becomes `-\int(1-t²)t⁴ dt`. We multiply this out to `-(t⁴-t⁶)` and integrate each part. At the end, we put `\cos x` back in place of `t`.

🎯 Exam Tip: When integrating powers of `\sin x` and `\cos x` where one power is odd, save one factor of the odd power's function (e.g., `\sin x`), convert the remaining even powers to the other function using `\sin² x + \cos² x = 1`, and then use substitution. For example, if `\sin x` has an odd power, use `t = \cos x`.