GSEB Class 12 Maths Solutions Chapter 7 સંકલન Exercise 7.2

નીચેના વિધેયોનાં સંકલિત મેળવો : (1 થી 37)

Question 1. \( \frac{2x}{1+x^2} \)
Answer:
Let \( I = \int \frac{2x}{1+x^2} dx \)
Suppose \( 1+x^2 = t \)
Differentiating both sides, we get \( 2x \, dx = dt \)
Then \( I = \int \frac{1}{t} dt \)
\( = \log |t| + c \)
\( = \log |1+x^2| + c \) (since \( t = 1+x^2 \))
In simple words: To integrate this, we substitute \( 1+x^2 \) with \( t \). Then we differentiate to find \( dt \). After substitution, the integral becomes a simple logarithm. Finally, we replace \( t \) with its original expression.

Exam Tip: For rational functions where the numerator is the derivative of the denominator, remember the direct integration formula \( \int \frac{f'(x)}{f(x)} dx = \log |f(x)| + C \). This saves time.

Question 2. \( \frac{(\log x)^2}{x} \)
Answer:
Let \( I = \int \frac{(\log x)^2}{x} dx \)
Suppose \( \log x = t \)
Differentiating both sides, we get \( \frac{1}{x} dx = dt \)
Then \( I = \int t^2 dt \)
\( = \frac{t^3}{3} + c \)
\( = \frac{(\log x)^3}{3} + c \)
In simple words: We let \( \log x \) be \( t \). After finding \( dt \), the integral becomes a simple power function of \( t \). We then integrate that and substitute \( t \) back with \( \log x \) for the final result.

Exam Tip: When you see \( \log x \) and \( \frac{1}{x} \) in the same integral, it's a strong hint to substitute \( \log x = t \).

Question 3. \( \frac{1}{x+x \log x} \)
Answer:
Let \( I = \int \frac{1}{x+x \log x} dx \)
\( = \int \frac{1}{x(1+\log x)} dx \)
Suppose \( 1+\log x = t \)
Differentiating both sides, we get \( \frac{1}{x} dx = dt \)
Then \( I = \int \frac{1}{t} dt \)
\( = \log |t| + c \)
\( = \log |1+\log x| + c \) (since \( t = 1+\log x \))
In simple words: First, factor out \( x \) from the denominator. Then, let \( 1+\log x \) be \( t \). After differentiating to find \( dt \), the integral simplifies to \( \frac{1}{t} \), which integrates to \( \log|t| \). Replace \( t \) to get the final answer.

Exam Tip: Always look for common factors in the denominator before attempting substitution, as it often simplifies the expression for easier integration.

Question 4. \( \sin x \sin(\cos x) \)
Answer:
Let \( I = \int \sin x \sin(\cos x) dx \)
Suppose \( \cos x = t \)
Differentiating both sides, we get \( -\sin x \, dx = dt \)
\( \implies \sin x \, dx = -dt \)
Then \( I = \int \sin t (-dt) \)
\( = -\int \sin t \, dt \)
\( = -(-\cos t) + c \)
\( = \cos t + c \)
\( = \cos(\cos x) + c \) (since \( t = \cos x \))
In simple words: We take \( \cos x \) as \( t \). When we differentiate, we get \( -\sin x \, dx \), so \( \sin x \, dx \) becomes \( -dt \). This turns the integral into \( -\int \sin t \, dt \), which integrates to \( \cos t \). Finally, we put \( \cos x \) back in place of \( t \).

Exam Tip: When dealing with composite trigonometric functions, try substituting the "inner" function. The derivative of the inner function often appears elsewhere in the integrand.

Question 5. \( \sin (ax + b) \cos (ax + b) \)
Answer:
Let \( I = \int \sin (ax + b) \cos (ax + b) dx \)
We can rewrite the integrand using the identity \( 2 \sin \theta \cos \theta = \sin (2\theta) \).
So, \( I = \int \frac{1}{2} \cdot 2 \sin (ax + b) \cos (ax + b) dx \)
\( = \frac{1}{2} \int \sin (2(ax + b)) dx \)
\( = \frac{1}{2} \int \sin (2ax + 2b) dx \)
Suppose \( 2ax + 2b = t \)
Then \( 2a \, dx = dt \)
\( \implies dx = \frac{dt}{2a} \)
So, \( I = \frac{1}{2} \int \sin t \cdot \frac{dt}{2a} \)
\( = \frac{1}{4a} \int \sin t \, dt \)
\( = \frac{1}{4a} (-\cos t) + c \)
\( = -\frac{1}{4a} \cos (2ax + 2b) + c \) (since \( t = 2ax + 2b \))
\( = -\frac{1}{4a} \cos 2(ax + b) + c \)
In simple words: We use a trigonometric identity to change \( \sin \theta \cos \theta \) into \( \frac{1}{2} \sin(2\theta) \). Then, we substitute \( 2ax+2b \) with \( t \) to simplify the integral. After integrating \( \sin t \), we replace \( t \) to get the final function.

Exam Tip: Always look for trigonometric identities that can simplify the integrand before applying substitution. This often leads to a more straightforward integration path.

Question 6. \( \sqrt{ax+b} \)
Answer:
Let \( I = \int \sqrt{ax+b} dx \)
Suppose \( ax+b = t \)
Then \( a \, dx = dt \)
\( \implies dx = \frac{dt}{a} \)
So, \( I = \int \sqrt{t} \cdot \frac{dt}{a} \)
\( = \frac{1}{a} \int t^{\frac{1}{2}} dt \)
\( = \frac{1}{a} \cdot \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1} + c \)
\( = \frac{1}{a} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + c \)
\( = \frac{1}{a} \cdot \frac{2}{3} t^{\frac{3}{2}} + c \)
\( = \frac{2}{3a} (ax+b)^{\frac{3}{2}} + c \) (since \( t = ax+b \))
In simple words: We substitute \( ax+b \) with \( t \) and find \( dx \) in terms of \( dt \). The integral then becomes a power of \( t \), which is easy to integrate. Finally, we change \( t \) back to \( ax+b \) for the answer.

Exam Tip: Remember the power rule for integration, \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \) (for \( n \neq -1 \)). Apply it carefully when dealing with fractional powers from square roots.

Question 7. \( x\sqrt{x+2} \)
Answer:
Let \( I = \int x\sqrt{x+2} dx \)
Suppose \( x+2 = t^2 \)
Then \( x = t^2 - 2 \)
And \( dx = 2t \, dt \)
So, \( I = \int (t^2 - 2) \sqrt{t^2} (2t \, dt) \)
\( = \int (t^2 - 2) t (2t \, dt) \)
\( = \int (t^2 - 2) (2t^2 \, dt) \)
\( = \int (2t^4 - 4t^2) dt \)
\( = 2 \int t^4 dt - 4 \int t^2 dt \)
\( = 2 \frac{t^5}{5} - 4 \frac{t^3}{3} + c \)
Since \( t = \sqrt{x+2} = (x+2)^{\frac{1}{2}} \), we substitute this back:
\( = \frac{2}{5} (x+2)^{\frac{5}{2}} - \frac{4}{3} (x+2)^{\frac{3}{2}} + c \)
In simple words: To integrate this, we let \( x+2 = t^2 \) which helps us get rid of the square root. We also express \( x \) and \( dx \) in terms of \( t \). This transforms the integral into a simpler polynomial of \( t \). After integrating term by term, we substitute \( t \) back with \( \sqrt{x+2} \) to get the final result.

Exam Tip: When you have an \( x \) term outside a square root like \( \sqrt{ax+b} \), it's often useful to substitute the entire expression under the root with \( t^2 \) to simplify both the root and the \( x \) term.

Question 8. \( x\sqrt{1+2 x^2} \)
Answer:
Let \( I = \int x\sqrt{1+2x^2} dx \)
Suppose \( 1+2x^2 = t \)
Then \( 4x \, dx = dt \)
\( \implies x \, dx = \frac{dt}{4} \)
So, \( I = \int \sqrt{t} \cdot \frac{dt}{4} \)
\( = \frac{1}{4} \int t^{\frac{1}{2}} dt \)
\( = \frac{1}{4} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + c \)
\( = \frac{1}{4} \cdot \frac{2}{3} t^{\frac{3}{2}} + c \)
\( = \frac{1}{6} t^{\frac{3}{2}} + c \)
\( = \frac{1}{6} (1+2x^2)^{\frac{3}{2}} + c \) (since \( t = 1+2x^2 \))
In simple words: We consider \( 1+2x^2 \) as \( t \). Then we find the derivative \( 4x \, dx \), which means \( x \, dx \) is \( \frac{dt}{4} \). The integral then becomes \( \frac{1}{4} \int \sqrt{t} \, dt \), which we can easily solve using the power rule. Finally, we substitute \( t \) back to \( 1+2x^2 \).

Exam Tip: Always look for a part of the integrand whose derivative (or a constant multiple of it) is also present in the integral. This signals a good candidate for substitution.

Question 9. \( (4x + 2)\sqrt{x^2+x+1} \)
Answer:
Let \( I = \int (4x + 2)\sqrt{x^2+x+1} dx \)
\( = \int 2(2x + 1)\sqrt{x^2+x+1} dx \)
Suppose \( x^2+x+1 = t \)
Then \( (2x+1) \, dx = dt \)
So, \( I = \int 2\sqrt{t} \, dt \)
\( = 2 \int t^{\frac{1}{2}} dt \)
\( = 2 \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + c \)
\( = 2 \cdot \frac{2}{3} t^{\frac{3}{2}} + c \)
\( = \frac{4}{3} t^{\frac{3}{2}} + c \)
\( = \frac{4}{3} (x^2+x+1)^{\frac{3}{2}} + c \) (since \( t = x^2+x+1 \))
In simple words: First, we factor out 2 from \( (4x+2) \). Then, we let \( x^2+x+1 \) be \( t \), which means \( (2x+1)dx \) is \( dt \). The integral becomes \( \int 2\sqrt{t} \, dt \), which simplifies nicely. After integrating, we replace \( t \) with its original quadratic expression.

Exam Tip: Recognize when an expression outside a root is a multiple of the derivative of the expression inside the root. This is a classic substitution problem.

Question 10. \( \frac{1}{x-\sqrt{x}} \)
Answer:
Let \( I = \int \frac{1}{x-\sqrt{x}} dx \)
Factor out \( \sqrt{x} \) from the denominator:
\( = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)} dx \)
Suppose \( \sqrt{x}-1 = t \)
Then \( \frac{1}{2\sqrt{x}} dx = dt \)
\( \implies \frac{1}{\sqrt{x}} dx = 2dt \)
So, \( I = \int \frac{1}{t} (2dt) \)
\( = 2 \int \frac{1}{t} dt \)
\( = 2 \log |t| + c \)
\( = 2 \log |\sqrt{x}-1| + c \) (since \( t = \sqrt{x}-1 \))
In simple words: First, simplify the denominator by factoring out \( \sqrt{x} \). Then, let \( \sqrt{x}-1 \) be \( t \). Find \( dt \) by differentiating. The integral then becomes \( 2\int \frac{1}{t} dt \), which integrates to \( 2\log|t| \). Substitute \( t \) back to get the final answer.

Exam Tip: When dealing with \( x \) and \( \sqrt{x} \) in the denominator, try factoring out \( \sqrt{x} \). This often reveals a clear path for substitution involving \( \sqrt{x} \).

Question 11. \( \frac{x}{\sqrt{x+4}} \), \( x > -4 \)
Answer:
Let \( I = \int \frac{x}{\sqrt{x+4}} dx \)
Suppose \( x+4 = t \)
Then \( x = t-4 \)
And \( dx = dt \)
So, \( I = \int \frac{t-4}{\sqrt{t}} dt \)
\( = \int \left( \frac{t}{\sqrt{t}} - \frac{4}{\sqrt{t}} \right) dt \)
\( = \int \left( t^{\frac{1}{2}} - 4t^{-\frac{1}{2}} \right) dt \)
\( = \int t^{\frac{1}{2}} dt - 4 \int t^{-\frac{1}{2}} dt \)
\( = \frac{t^{\frac{3}{2}}}{\frac{3}{2}} - 4 \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + c \)
\( = \frac{2}{3} t^{\frac{3}{2}} - 8 t^{\frac{1}{2}} + c \)
Substitute \( t = x+4 \) back:
\( = \frac{2}{3} (x+4)^{\frac{3}{2}} - 8 (x+4)^{\frac{1}{2}} + c \)
In simple words: We let \( x+4 = t \). This also means \( x \) is \( t-4 \) and \( dx \) is \( dt \). We substitute these into the integral, separate the terms, and integrate each one using the power rule. Finally, we replace \( t \) with \( x+4 \) to find the answer.

Exam Tip: When the numerator contains \( x \) and the denominator has \( \sqrt{ax+b} \), substituting \( t = ax+b \) (or \( t^2 = ax+b \)) allows you to express \( x \) in terms of \( t \) and simplify the integrand into separable terms.

Question 12. \( (x^3 - 1)^{\frac{1}{3}}x^5 \)
Answer:
Let \( I = \int (x^3 - 1)^{\frac{1}{3}}x^5 dx \)
Suppose \( x^3-1 = t \)
Then \( x^3 = t+1 \)
And \( 3x^2 \, dx = dt \)
\( \implies x^2 \, dx = \frac{dt}{3} \)
Now, rewrite \( x^5 \) as \( x^3 \cdot x^2 \):
\( I = \int (x^3 - 1)^{\frac{1}{3}} x^3 \cdot x^2 dx \)
Substitute \( (x^3-1) = t \), \( x^3 = t+1 \), and \( x^2 \, dx = \frac{dt}{3} \):
\( I = \int t^{\frac{1}{3}} (t+1) \frac{dt}{3} \)
\( = \frac{1}{3} \int (t^{\frac{1}{3}} \cdot t + t^{\frac{1}{3}} \cdot 1) dt \)
\( = \frac{1}{3} \int (t^{\frac{4}{3}} + t^{\frac{1}{3}}) dt \)
\( = \frac{1}{3} \left( \frac{t^{\frac{4}{3}+1}}{\frac{4}{3}+1} + \frac{t^{\frac{1}{3}+1}}{\frac{1}{3}+1} \right) + c \)
\( = \frac{1}{3} \left( \frac{t^{\frac{7}{3}}}{\frac{7}{3}} + \frac{t^{\frac{4}{3}}}{\frac{4}{3}} \right) + c \)
\( = \frac{1}{3} \left( \frac{3}{7} t^{\frac{7}{3}} + \frac{3}{4} t^{\frac{4}{3}} \right) + c \)
\( = \frac{1}{7} t^{\frac{7}{3}} + \frac{1}{4} t^{\frac{4}{3}} + c \)
Substitute \( t = x^3-1 \) back:
\( = \frac{1}{7} (x^3-1)^{\frac{7}{3}} + \frac{1}{4} (x^3-1)^{\frac{4}{3}} + c \)
In simple words: We let \( x^3-1 \) be \( t \), which means \( x^3 \) becomes \( t+1 \). We also find \( x^2 \, dx \) in terms of \( dt \). By splitting \( x^5 \) into \( x^3 \cdot x^2 \), we can substitute all terms with \( t \). This changes the integral into a simpler polynomial of \( t \) with fractional powers. After integrating, we replace \( t \) with \( x^3-1 \).

Exam Tip: When an integral has a term like \( (f(x))^n \cdot x^m \) where \( m \) is greater than the degree of \( f(x) \), try to split \( x^m \) into factors that can be used for substitution and factors that can be expressed in terms of \( t \).

Question 13. \( \frac{x^2}{(2+3 x^3)^3} \)
Answer:
Let \( I = \int \frac{x^2}{(2+3x^3)^3} dx \)
Suppose \( 2+3x^3 = t \)
Then \( 9x^2 \, dx = dt \)
\( \implies x^2 \, dx = \frac{dt}{9} \)
So, \( I = \int \frac{1}{t^3} \frac{dt}{9} \)
\( = \frac{1}{9} \int t^{-3} dt \)
\( = \frac{1}{9} \frac{t^{-3+1}}{-3+1} + c \)
\( = \frac{1}{9} \frac{t^{-2}}{-2} + c \)
\( = -\frac{1}{18} t^{-2} + c \)
\( = -\frac{1}{18t^2} + c \)
Substitute \( t = 2+3x^3 \) back:
\( = -\frac{1}{18(2+3x^3)^2} + c \)
In simple words: We set \( 2+3x^3 \) equal to \( t \). Then we differentiate to find \( 9x^2 \, dx = dt \), which gives us \( x^2 \, dx = \frac{dt}{9} \). Substituting these into the integral converts it to a power of \( t \), which is simple to integrate. Finally, we replace \( t \) with its original expression.

Exam Tip: For functions of the form \( \frac{f'(x)}{(f(x))^n} \), substitution \( t = f(x) \) simplifies the integral greatly to \( \int t^{-n} dt \).

Question 14. \( \frac{1}{x(\log x)^m}, x > 0, m \neq 1 \)
Answer:
Let \( I = \int \frac{1}{x(\log x)^m} dx \)
Suppose \( \log x = t \)
Then \( \frac{1}{x} dx = dt \)
So, \( I = \int \frac{1}{t^m} dt \)
\( = \int t^{-m} dt \)
\( = \frac{t^{-m+1}}{-m+1} + c \)
Substitute \( t = \log x \) back:
\( = \frac{(\log x)^{-m+1}}{-m+1} + c \)
In simple words: We let \( \log x \) be \( t \). Then \( \frac{1}{x} dx \) becomes \( dt \). The integral changes to \( \int t^{-m} dt \), which we can solve using the power rule. Finally, we replace \( t \) with \( \log x \) to get the solution.

Exam Tip: The derivative of \( \log x \) is \( \frac{1}{x} \). If both are present in the integrand, \( \log x = t \) is typically the correct substitution.

Question 15. \( \frac{x}{9-4x^2} \)
Answer:
Let \( I = \int \frac{x}{9-4x^2} dx \)
Suppose \( 9-4x^2 = t \)
Then \( -8x \, dx = dt \)
\( \implies x \, dx = \frac{dt}{-8} \)
So, \( I = \int \frac{1}{t} \left( \frac{dt}{-8} \right) \)
\( = -\frac{1}{8} \int \frac{1}{t} dt \)
\( = -\frac{1}{8} \log |t| + c \)
Substitute \( t = 9-4x^2 \) back:
\( = -\frac{1}{8} \log |9-4x^2| + c \)
In simple words: We make a substitution \( 9-4x^2 = t \). When we differentiate, we get \( -8x \, dx = dt \), so \( x \, dx \) becomes \( -\frac{1}{8} dt \). This converts the integral into \( -\frac{1}{8} \int \frac{1}{t} dt \), which integrates to a logarithm. Finally, we replace \( t \) with its original expression.

Exam Tip: Remember to handle constants carefully during substitution. If \( du = k \, dx \), then \( dx = \frac{du}{k} \), and the constant \( \frac{1}{k} \) must be pulled outside the integral.

Question 16. \( e^{2x+3} \)
Answer:
Let \( I = \int e^{2x+3} dx \)
Suppose \( 2x+3 = t \)
Then \( 2 \, dx = dt \)
\( \implies dx = \frac{dt}{2} \)
So, \( I = \int e^t \frac{dt}{2} \)
\( = \frac{1}{2} \int e^t dt \)
\( = \frac{1}{2} e^t + c \)
Substitute \( t = 2x+3 \) back:
\( = \frac{1}{2} e^{2x+3} + c \)
In simple words: We let the exponent \( 2x+3 \) be \( t \). Differentiating gives \( 2 \, dx = dt \), so \( dx \) becomes \( \frac{dt}{2} \). The integral simplifies to \( \frac{1}{2} \int e^t dt \). After integration, we substitute \( t \) back with \( 2x+3 \).

Exam Tip: For integrals of the form \( \int e^{ax+b} dx \), the result is \( \frac{1}{a} e^{ax+b} + C \). Knowing this pattern can speed up solving such problems.

Question 17. \( \frac{x}{e^{x^2}} \)
Answer:
Let \( I = \int \frac{x}{e^{x^2}} dx \)
\( = \int x e^{-x^2} dx \)
Suppose \( -x^2 = t \)
Then \( -2x \, dx = dt \)
\( \implies x \, dx = -\frac{dt}{2} \)
So, \( I = \int e^t \left(-\frac{dt}{2}\right) \)
\( = -\frac{1}{2} \int e^t dt \)
\( = -\frac{1}{2} e^t + c \)
Substitute \( t = -x^2 \) back:
\( = -\frac{1}{2} e^{-x^2} + c \)
\( = -\frac{1}{2e^{x^2}} + c \)
In simple words: We let the exponent \( -x^2 \) be \( t \). Then, \( x \, dx \) turns into \( -\frac{1}{2} dt \). The integral becomes \( -\frac{1}{2} \int e^t dt \), which is straightforward to integrate. Finally, we put \( -x^2 \) back in place of \( t \).

Exam Tip: When \( x \) is in the numerator and \( e^{x^2} \) (or a similar form) is in the denominator, consider substituting the exponent of \( e \). The derivative of \( x^2 \) will provide the \( x \, dx \) term needed for substitution.

Question 18. \( \frac{e^{\tan^{-1} x}}{1+x^2} \)
Answer:
Let \( I = \int \frac{e^{\tan^{-1} x}}{1+x^2} dx \)
Suppose \( \tan^{-1} x = t \)
Then \( \frac{1}{1+x^2} dx = dt \)
So, \( I = \int e^t dt \)
\( = e^t + c \)
Substitute \( t = \tan^{-1} x \) back:
\( = e^{\tan^{-1} x} + c \)
In simple words: We substitute \( \tan^{-1} x \) with \( t \). The derivative of \( \tan^{-1} x \) is \( \frac{1}{1+x^2} \), which is also in the integral, becoming \( dt \). So, the integral becomes \( \int e^t dt \), which is \( e^t \). Finally, we put \( \tan^{-1} x \) back for \( t \).

Exam Tip: Always recognize the derivatives of inverse trigonometric functions. The presence of \( \frac{1}{1+x^2} \) strongly suggests a substitution involving \( \tan^{-1} x \).

Question 19. \( \frac{e^{2x}-1}{e^{2x}+1} \)
Answer:
Let \( I = \int \frac{e^{2x}-1}{e^{2x}+1} dx \)
Divide the numerator and denominator by \( e^x \):
\( = \int \frac{\frac{e^{2x}}{e^x} - \frac{1}{e^x}}{\frac{e^{2x}}{e^x} + \frac{1}{e^x}} dx \)
\( = \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx \)
Suppose \( e^x + e^{-x} = t \)
Then \( (e^x - e^{-x}) \, dx = dt \)
So, \( I = \int \frac{1}{t} dt \)
\( = \log |t| + c \)
Substitute \( t = e^x + e^{-x} \) back:
\( = \log |e^x + e^{-x}| + c \)
In simple words: First, we divide the top and bottom of the fraction by \( e^x \) to simplify the expression. Then, we let the denominator, \( e^x + e^{-x} \), be \( t \). The derivative of this (the numerator) becomes \( dt \). This leaves us with \( \int \frac{1}{t} dt \), which is \( \log|t| \). Finally, we replace \( t \) with \( e^x + e^{-x} \).

Exam Tip: For rational functions involving exponential terms like \( e^{2x} \), try dividing the numerator and denominator by \( e^x \) or \( e^{-x} \) to create a form where the numerator is the derivative of the denominator.

Question 20. \( \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} \)
Answer:
Let \( I = \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} dx \)
Suppose \( e^{2x}+e^{-2x} = t \)
Then \( (2e^{2x} - 2e^{-2x}) \, dx = dt \)
\( \implies 2(e^{2x} - e^{-2x}) \, dx = dt \)
\( \implies (e^{2x} - e^{-2x}) \, dx = \frac{dt}{2} \)
So, \( I = \int \frac{1}{t} \frac{dt}{2} \)
\( = \frac{1}{2} \int \frac{1}{t} dt \)
\( = \frac{1}{2} \log |t| + c \)
Substitute \( t = e^{2x}+e^{-2x} \) back:
\( = \frac{1}{2} \log |e^{2x}+e^{-2x}| + c \)
In simple words: We let the denominator \( e^{2x}+e^{-2x} \) be \( t \). When we differentiate, the numerator \( e^{2x}-e^{-2x} \) (multiplied by a constant) becomes part of \( dt \). The integral simplifies to \( \frac{1}{2} \int \frac{1}{t} dt \), which is \( \frac{1}{2} \log|t| \). We then replace \( t \) with its original exponential expression.

Exam Tip: This 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 of the denominator (or a constant multiple of it).

Question 21. \( \tan^2(2x-3) \)
Answer:
Let \( I = \int \tan^2(2x-3) dx \)
Use the identity \( \tan^2 \theta = \sec^2 \theta - 1 \):
\( I = \int (\sec^2(2x-3) - 1) dx \)
\( = \int \sec^2(2x-3) dx - \int 1 \, dx \)
For the first integral, let \( 2x-3 = t \)
Then \( 2 \, dx = dt \)
\( \implies dx = \frac{dt}{2} \)
So, \( \int \sec^2(2x-3) dx = \int \sec^2 t \frac{dt}{2} \)
\( = \frac{1}{2} \int \sec^2 t dt \)
\( = \frac{1}{2} \tan t + C_1 \)
\( = \frac{1}{2} \tan(2x-3) + C_1 \)
The second integral is \( \int 1 \, dx = x + C_2 \)
Combining them: \( I = \frac{1}{2} \tan(2x-3) - x + c \)
where \( c = C_1 - C_2 \) is the constant of integration.
In simple words: First, we use the identity \( \tan^2 \theta = \sec^2 \theta - 1 \) to change the integral. Then we split it into two simpler integrals. For the \( \sec^2 \) part, we substitute \( 2x-3 \) with \( t \) to solve it. The second part is just \( \int 1 \, dx \). We combine the results for the final answer.

Exam Tip: Always remember the fundamental trigonometric identities like \( \tan^2 \theta = \sec^2 \theta - 1 \) and \( \cot^2 \theta = \csc^2 \theta - 1 \), as they are essential for integrating squared tangent and cotangent functions.

Question 22. \( \sec^2(7-4x) \)
Answer:
Let \( I = \int \sec^2(7-4x) dx \)
Suppose \( 7-4x = t \)
Then \( -4 \, dx = dt \)
\( \implies dx = -\frac{dt}{4} \)
So, \( I = \int \sec^2 t \left(-\frac{dt}{4}\right) \)
\( = -\frac{1}{4} \int \sec^2 t dt \)
\( = -\frac{1}{4} \tan t + c \)
Substitute \( t = 7-4x \) back:
\( = -\frac{1}{4} \tan(7-4x) + c \)
In simple words: We let \( 7-4x \) be \( t \). When we differentiate, \( -4 \, dx = dt \), so \( dx \) becomes \( -\frac{1}{4} dt \). The integral simplifies to \( -\frac{1}{4} \int \sec^2 t dt \), which integrates to \( \tan t \). Finally, we put \( 7-4x \) back in place of \( t \).

Exam Tip: The integral of \( \sec^2 \theta \) is \( \tan \theta \). If the argument of \( \sec^2 \) is linear, like \( ax+b \), the integral is \( \frac{1}{a} \tan(ax+b) + C \). Be careful with the sign if \( a \) is negative.

Question 23. \( \frac{\sin^{-1} x}{\sqrt{1-x^2}} \)
Answer:
Let \( I = \int \frac{\sin^{-1} x}{\sqrt{1-x^2}} dx \)
Suppose \( \sin^{-1} x = t \)
Then \( \frac{1}{\sqrt{1-x^2}} dx = dt \)
So, \( I = \int t \, dt \)
\( = \frac{t^2}{2} + c \)
Substitute \( t = \sin^{-1} x \) back:
\( = \frac{(\sin^{-1} x)^2}{2} + c \)
In simple words: We take \( \sin^{-1} x \) as \( t \). Its derivative, \( \frac{1}{\sqrt{1-x^2}} dx \), is also present in the integral, making it \( dt \). The integral then becomes \( \int t \, dt \), which integrates to \( \frac{t^2}{2} \). Finally, we put \( \sin^{-1} x \) back for \( t \).

Exam Tip: Recognize the derivative of \( \sin^{-1} x \). If you see \( \sin^{-1} x \) multiplied by \( \frac{1}{\sqrt{1-x^2}} \), it's a strong indicator to use \( \sin^{-1} x = t \) as a substitution.

Question 24. \( \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} \)
Answer:
Let \( I = \int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} dx \)
The denominator can be written as \( 2(3 \cos x + 2 \sin x) \).
So, \( I = \int \frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)} dx \)
Suppose \( 3 \cos x + 2 \sin x = t \)
Then \( (-3 \sin x + 2 \cos x) \, dx = dt \)
So, \( (2 \cos x - 3 \sin x) \, dx = dt \)
Substitute these into the integral:
\( I = \int \frac{1}{2t} dt \)
\( = \frac{1}{2} \int \frac{1}{t} dt \)
\( = \frac{1}{2} \log |t| + c \)
Substitute \( t = 3 \cos x + 2 \sin x \) back:
\( = \frac{1}{2} \log |3 \cos x + 2 \sin x| + c \)
In simple words: First, we factor out 2 from the denominator. Then, we let the expression \( 3 \cos x + 2 \sin x \) be \( t \). Its derivative, \( (2 \cos x - 3 \sin x) \, dx \), becomes \( dt \). The integral simplifies to \( \frac{1}{2} \int \frac{1}{t} dt \), which integrates to \( \frac{1}{2} \log|t| \). Finally, we replace \( t \) with its original trigonometric expression.

Exam Tip: For rational functions of trigonometric expressions, check if the numerator is the derivative of the denominator (or a constant multiple of it). Factoring constants out of the denominator can often make this relationship clear.

Question 25. Integrate: \( \frac{1}{\cos ^2 x(1-\tan x)^2} \)
Answer:Let \( I = \int \frac{1}{\cos^2 x (1-\tan x)^2} \, dx \)
This can be rewritten as \( I = \int \frac{\sec^2 x}{(1-\tan x)^2} \, dx \)
Let \( 1 - \tan x = t \).
Differentiating both sides, we get \( -\sec^2 x \, dx = dt \).
This implies \( \sec^2 x \, dx = -dt \).
Substitute these into the integral:
\( I = \int \frac{-dt}{t^2} \)
\( I = - \int t^{-2} \, dt \)
Using the power rule for integration \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \), we have:
\( I = - \left( \frac{t^{-2+1}}{-2+1} \right) + c \)
\( I = - \left( \frac{t^{-1}}{-1} \right) + c \)
\( I = \frac{1}{t} + c \)
Substitute back \( t = 1 - \tan x \):
\( I = \frac{1}{1 - \tan x} + c \)
In simple words: To solve this, we first change the fraction to use secant. Then, we let a part of the expression be 't' and find its derivative. Replacing these in the integral allows us to solve it with a basic power rule, then we switch 't' back to the original expression.

Exam Tip: For integrals involving trigonometric functions and their derivatives, substitution is a key method. Always look for a part of the integrand whose derivative is also present (or a multiple of it).

Question 26. Integrate: \( \frac{\cos \sqrt{x}}{\sqrt{x}} \)
Answer:Let \( I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} \, dx \)
Let \( \sqrt{x} = t \).
Differentiating both sides with respect to \( x \):
\( \frac{1}{2\sqrt{x}} \, dx = dt \)
This means \( \frac{1}{\sqrt{x}} \, dx = 2 \, dt \).
Substitute these into the integral:
\( I = \int \cos t \cdot (2 \, dt) \)
\( I = 2 \int \cos t \, dt \)
The integral of \( \cos t \) is \( \sin t \).
\( I = 2 \sin t + c \)
Substitute back \( t = \sqrt{x} \):
\( I = 2 \sin \sqrt{x} + c \)
In simple words: We simplify this integral by substituting \( \sqrt{x} \) with 't'. After finding the derivative of 't', we replace the terms in the integral. This turns it into a simpler cosine integral, which we then solve and substitute 't' back to get the final answer.

Exam Tip: When \( \sqrt{x} \) appears in an integral, it is often useful to substitute \( t = \sqrt{x} \) because its derivative \( \frac{1}{2\sqrt{x}} \) is also simple to work with.

Question 27. Integrate: \( \sqrt{\sin 2 x} \cos 2x \)
Answer:Let \( I = \int \sqrt{\sin 2x} \cos 2x \, dx \)
Let \( \sin 2x = t \).
Differentiating both sides with respect to \( x \):
\( 2 \cos 2x \, dx = dt \)
This implies \( \cos 2x \, dx = \frac{dt}{2} \).
Substitute these into the integral:
\( I = \int \sqrt{t} \cdot \frac{dt}{2} \)
\( I = \frac{1}{2} \int t^{1/2} \, dt \)
Using the power rule for integration:
\( I = \frac{1}{2} \left( \frac{t^{1/2+1}}{1/2+1} \right) + c \)
\( I = \frac{1}{2} \left( \frac{t^{3/2}}{3/2} \right) + c \)
\( I = \frac{1}{2} \cdot \frac{2}{3} t^{3/2} + c \)
\( I = \frac{1}{3} t^{3/2} + c \)
Substitute back \( t = \sin 2x \):
\( I = \frac{1}{3} (\sin 2x)^{3/2} + c \)
In simple words: For this integral, we substitute \( \sin 2x \) with 't'. After calculating the derivative of 't', we replace it in the integral. This changes the problem to a simple power rule integration, which we then solve and then substitute 't' back into the result.

Exam Tip: When a function and its derivative (or a multiple of its derivative) are present in an integral, a u-substitution (or t-substitution) is almost always the correct approach.

Question 28. Integrate: \( \frac{\cos x}{\sqrt{1+\sin x}} \)
Answer:Let \( I = \int \frac{\cos x}{\sqrt{1+\sin x}} \, dx \)
Let \( 1 + \sin x = t \).
Differentiating both sides with respect to \( x \):
\( \cos x \, dx = dt \).
Substitute these into the integral:
\( I = \int \frac{dt}{\sqrt{t}} \)
\( I = \int t^{-1/2} \, dt \)
Using the power rule for integration:
\( I = \frac{t^{-1/2+1}}{-1/2+1} + c \)
\( I = \frac{t^{1/2}}{1/2} + c \)
\( I = 2t^{1/2} + c \)
\( I = 2\sqrt{t} + c \)
Substitute back \( t = 1 + \sin x \):
\( I = 2\sqrt{1+\sin x} + c \)
In simple words: We tackle this integral by letting the expression inside the square root be 't'. After finding its derivative, we replace the terms in the integral. This simplifies it to a power rule integration, which we solve and then substitute 't' back into the final answer.

Exam Tip: When you see \( 1 + \sin x \) under a square root and \( \cos x \) elsewhere, it's a strong hint to use substitution with \( t = 1 + \sin x \).

Question 29. Integrate: \( \cot x \log \sin x \)
Answer:Let \( I = \int \cot x \log \sin x \, dx \)
Let \( \log \sin x = t \).
Differentiating both sides with respect to \( x \):
\( \frac{1}{\sin x} \cdot \cos x \, dx = dt \)
\( \cot x \, dx = dt \).
Substitute these into the integral:
\( I = \int t \, dt \)
Using the power rule for integration:
\( I = \frac{t^2}{2} + c \)
Substitute back \( t = \log \sin x \):
\( I = \frac{(\log \sin x)^2}{2} + c \)
In simple words: To integrate this, we use a substitution where 't' is equal to \( \log \sin x \). By differentiating 't', we find that \( \cot x \, dx \) becomes 'dt'. This simplifies the integral to a basic power form, which we then solve and put 't' back into the result.

Exam Tip: Recognize that \( \cot x \) is the derivative of \( \log \sin x \). This makes \( t = \log \sin x \) an effective substitution for problems like this.

Question 30. Integrate: \( \frac{\sin x}{1+\cos x} \)
Answer:Let \( I = \int \frac{\sin x}{1+\cos x} \, dx \)
Let \( 1 + \cos x = t \).
Differentiating both sides with respect to \( x \):
\( -\sin x \, dx = dt \)
This means \( \sin x \, dx = -dt \).
Substitute these into the integral:
\( I = \int \frac{-dt}{t} \)
\( I = - \int \frac{1}{t} \, dt \)
The integral of \( \frac{1}{t} \) is \( \log |t| \).
\( I = - \log |t| + c \)
Substitute back \( t = 1 + \cos x \):
\( I = - \log |1 + \cos x| + c \)
In simple words: We find the integral by setting the denominator, \( 1 + \cos x \), equal to 't'. Then we differentiate 't' to find \( \sin x \, dx \). Substituting these into the integral makes it simpler, allowing us to use the log rule for integration. Finally, we put 't' back in.

Exam Tip: When the numerator is the derivative (or a multiple of the derivative) of the denominator, the integral will often involve a logarithm: \( \int \frac{f'(x)}{f(x)} \, dx = \log |f(x)| + C \).

Question 31. Integrate: \( \frac{\sin x}{(1+\cos x)^2} \)
Answer:Let \( I = \int \frac{\sin x}{(1+\cos x)^2} \, dx \)
Let \( 1 + \cos x = t \).
Differentiating both sides with respect to \( x \):
\( -\sin x \, dx = dt \)
This implies \( \sin x \, dx = -dt \).
Substitute these into the integral:
\( I = \int \frac{-dt}{t^2} \)
\( I = - \int t^{-2} \, dt \)
Using the power rule for integration:
\( I = - \left( \frac{t^{-2+1}}{-2+1} \right) + c \)
\( I = - \left( \frac{t^{-1}}{-1} \right) + c \)
\( I = \frac{1}{t} + c \)
Substitute back \( t = 1 + \cos x \):
\( I = \frac{1}{1 + \cos x} + c \)
In simple words: We solve this integral by letting \( 1 + \cos x \) be 't'. Then, we find the derivative of 't' which gives us \( -\sin x \, dx = dt \). Substituting these into the integral makes it easier to solve using the power rule for integration. Finally, we replace 't' with the original expression.

Exam Tip: When you see a function and its derivative (or a multiple of it) in a fraction, and the function is in the denominator with a power, a substitution of the denominator (or part of it) to 't' is usually effective.

Question 32. Integrate: \( \frac{1}{1+\cot x} \)
Answer:Let \( I = \int \frac{1}{1+\cot x} \, dx \)
Rewrite \( \cot x \) as \( \frac{\cos x}{\sin x} \):
\( I = \int \frac{1}{1+\frac{\cos x}{\sin x}} \, dx \)
\( I = \int \frac{1}{\frac{\sin x + \cos x}{\sin x}} \, dx \)
\( I = \int \frac{\sin x}{\sin x + \cos x} \, dx \)
To simplify, multiply the numerator and denominator by 2:
\( I = \frac{1}{2} \int \frac{2\sin x}{\sin x + \cos x} \, dx \)
Rewrite \( 2\sin x \) as \( (\sin x + \cos x) - (\cos x - \sin x) \):
\( I = \frac{1}{2} \int \frac{(\sin x + \cos x) - (\cos x - \sin x)}{\sin x + \cos x} \, dx \)
Separate the terms:
\( I = \frac{1}{2} \int \left( \frac{\sin x + \cos x}{\sin x + \cos x} - \frac{\cos x - \sin x}{\sin x + \cos x} \right) \, dx \)
\( I = \frac{1}{2} \int \left( 1 - \frac{\cos x - \sin x}{\sin x + \cos x} \right) \, dx \)
\( I = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \frac{\cos x - \sin x}{\sin x + \cos x} \, dx \)
The first integral is \( \frac{1}{2}x \).
For the second integral, let \( t = \sin x + \cos x \).
Differentiating, \( dt = (\cos x - \sin x) \, dx \).
So, \( \int \frac{\cos x - \sin x}{\sin x + \cos x} \, dx = \int \frac{dt}{t} = \log |t| + C_1 = \log |\sin x + \cos x| + C_1 \).
Combining everything:
\( I = \frac{1}{2}x - \frac{1}{2} \log |\sin x + \cos x| + c \)
In simple words: First, we change \( \cot x \) to \( \frac{\cos x}{\sin x} \) and simplify the fraction. Then, we cleverly split the numerator to create two terms: one that integrates to 'x' and another that matches the derivative of the denominator, allowing us to use a logarithm for the second part.

Exam Tip: For integrals of the form \( \int \frac{1}{1+\tan x} \, dx \) or \( \int \frac{1}{1+\cot x} \, dx \), it is often useful to rewrite the tangent/cotangent in terms of sine and cosine and then manipulate the numerator to create terms that are easily integrable (like the denominator or its derivative).

Question 33. Integrate: \( \frac{1}{1-\tan x} \)
Answer:Let \( I = \int \frac{1}{1-\tan x} \, dx \)
Rewrite \( \tan x \) as \( \frac{\sin x}{\cos x} \):
\( I = \int \frac{1}{1-\frac{\sin x}{\cos x}} \, dx \)
\( I = \int \frac{1}{\frac{\cos x - \sin x}{\cos x}} \, dx \)
\( I = \int \frac{\cos x}{\cos x - \sin x} \, dx \)
To simplify, multiply the numerator and denominator by 2:
\( I = \frac{1}{2} \int \frac{2\cos x}{\cos x - \sin x} \, dx \)
Rewrite \( 2\cos x \) as \( (\cos x - \sin x) + (\cos x + \sin x) \):
\( I = \frac{1}{2} \int \frac{(\cos x - \sin x) + (\cos x + \sin x)}{\cos x - \sin x} \, dx \)
Separate the terms:
\( I = \frac{1}{2} \int \left( \frac{\cos x - \sin x}{\cos x - \sin x} + \frac{\cos x + \sin x}{\cos x - \sin x} \right) \, dx \)
\( I = \frac{1}{2} \int \left( 1 + \frac{\cos x + \sin x}{\cos x - \sin x} \right) \, dx \)
\( I = \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \frac{\cos x + \sin x}{\cos x - \sin x} \, dx \)
The first integral is \( \frac{1}{2}x \).
For the second integral, let \( t = \cos x - \sin x \).
Differentiating, \( dt = (-\sin x - \cos x) \, dx = -(\sin x + \cos x) \, dx \).
So, \( \int \frac{\cos x + \sin x}{\cos x - \sin x} \, dx = \int \frac{-dt}{t} = - \log |t| + C_1 = - \log |\cos x - \sin x| + C_1 \).
Combining everything:
\( I = \frac{1}{2}x - \frac{1}{2} \log |\cos x - \sin x| + c \)
In simple words: We begin by changing \( \tan x \) to \( \frac{\sin x}{\cos x} \) and simplifying the fraction. Then, we manipulate the numerator to create a sum of the denominator and its derivative. This lets us break the integral into two simpler parts: one that integrates to 'x' and another that involves a logarithm due to substitution.

Exam Tip: For integrals of the form \( \int \frac{1}{1-\tan x} \, dx \), transform \( \tan x \) to sine and cosine, and then use the trick \( 2\cos x = (\cos x - \sin x) + (\cos x + \sin x) \) in the numerator to simplify integration.

Question 34. Integrate: \( \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} \)
Answer:Let \( I = \int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} \, dx \)
To create \( \tan x \) and \( \sec^2 x \), divide the denominator by \( \cos x \) and multiply by \( \cos x \):
\( I = \int \frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x} \cdot \cos^2 x} \, dx \)
\( I = \int \frac{\sqrt{\tan x}}{\tan x \cdot \cos^2 x} \, dx \)
\( I = \int \frac{1}{\sqrt{\tan x}} \cdot \frac{1}{\cos^2 x} \, dx \)
\( I = \int \frac{1}{\sqrt{\tan x}} \cdot \sec^2 x \, dx \)
Let \( \tan x = t \).
Differentiating both sides with respect to \( x \):
\( \sec^2 x \, dx = dt \).
Substitute these into the integral:
\( I = \int \frac{1}{\sqrt{t}} \, dt \)
\( I = \int t^{-1/2} \, dt \)
Using the power rule for integration:
\( I = \frac{t^{-1/2+1}}{-1/2+1} + c \)
\( I = \frac{t^{1/2}}{1/2} + c \)
\( I = 2t^{1/2} + c \)
\( I = 2\sqrt{t} + c \)
Substitute back \( t = \tan x \):
\( I = 2\sqrt{\tan x} + c \)
In simple words: We begin by changing the denominator to include \( \tan x \) and \( \sec^2 x \), which helps to simplify the expression. Then, we substitute \( \tan x \) with 't' and find its derivative. Replacing these in the integral converts it into a basic power rule integral, which we solve and then substitute 't' back.

Exam Tip: When you see \( \sqrt{\tan x} \) and \( \sin x \cos x \) in the denominator, try to transform \( \sin x \cos x \) into \( \tan x \) and \( \sec^2 x \) by dividing and multiplying by \( \cos^2 x \). This creates the perfect setup for substitution with \( t = \tan x \).

Question 35. Integrate: \( \frac{(1+\log x)^2}{x} \)
Answer:Let \( I = \int \frac{(1+\log x)^2}{x} \, dx \)
Let \( 1 + \log x = t \).
Differentiating both sides with respect to \( x \):
\( \frac{1}{x} \, dx = dt \).
Substitute these into the integral:
\( I = \int t^2 \, dt \)
Using the power rule for integration:
\( I = \frac{t^3}{3} + c \)
Substitute back \( t = 1 + \log x \):
\( I = \frac{(1 + \log x)^3}{3} + c \)
In simple words: To integrate this expression, we substitute \( 1 + \log x \) with 't'. We then find the derivative of 't', which simplifies to \( \frac{1}{x} \, dx = dt \). Replacing these into the integral makes it an easy power rule problem, which we solve and then substitute 't' back to get the final answer.

Exam Tip: Whenever \( \log x \) is present in an integral, and \( \frac{1}{x} \) also appears (or can be easily created), consider substituting \( t = \log x \) or \( t = \text{expression involving } \log x \).

Question 36. Integrate: \( \frac{(x+1)(x+\log x)^2}{x} \)
Answer:Let \( I = \int \frac{(x+1)(x+\log x)^2}{x} \, dx \)
Rewrite the fraction \( \frac{x+1}{x} \) as \( 1 + \frac{1}{x} \):
\( I = \int \left( 1 + \frac{1}{x} \right) (x+\log x)^2 \, dx \)
Let \( x + \log x = t \).
Differentiating both sides with respect to \( x \):
\( \left( 1 + \frac{1}{x} \right) \, dx = dt \).
Substitute these into the integral:
\( I = \int t^2 \, dt \)
Using the power rule for integration:
\( I = \frac{t^3}{3} + c \)
Substitute back \( t = x + \log x \):
\( I = \frac{(x + \log x)^3}{3} + c \)
In simple words: We start by rewriting the fraction to separate the \( x+\log x \) term. Then, we substitute \( x + \log x \) with 't'. Finding the derivative of 't' neatly matches the remaining part of the integral. This lets us solve it using the simple power rule and then substitute 't' back.

Exam Tip: For expressions like \( \frac{x+1}{x} \), separating it into \( 1 + \frac{1}{x} \) can reveal a hidden derivative for substitution. Always look for ways to simplify the integrand before applying complex integration techniques.

Question 37. Integrate: \( \frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8} \)
Answer:Let \( I = \int \frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8} \, dx \)
Let \( \tan^{-1} x^4 = t \).
Differentiating both sides with respect to \( x \):
Recall the derivative rule \( \frac{d}{dx} (\tan^{-1} u) = \frac{1}{1+u^2} \frac{du}{dx} \). Here \( u = x^4 \), so \( \frac{du}{dx} = 4x^3 \).
\( \frac{1}{1+(x^4)^2} \cdot 4x^3 \, dx = dt \)
\( \frac{4x^3}{1+x^8} \, dx = dt \)
This means \( \frac{x^3}{1+x^8} \, dx = \frac{dt}{4} \).
Substitute these into the integral:
\( I = \int \sin t \cdot \frac{dt}{4} \)
\( I = \frac{1}{4} \int \sin t \, dt \)
The integral of \( \sin t \) is \( -\cos t \).
\( I = \frac{1}{4} (-\cos t) + c \)
\( I = -\frac{1}{4} \cos t + c \)
Substitute back \( t = \tan^{-1} x^4 \):
\( I = -\frac{1}{4} \cos (\tan^{-1} x^4) + c \)
In simple words: To solve this, we substitute \( \tan^{-1} x^4 \) with 't'. We then find its derivative, which gives us \( \frac{x^3}{1+x^8} \, dx = \frac{dt}{4} \). Substituting these into the integral simplifies it to a basic sine integral. We solve this and then replace 't' with its original expression.

Exam Tip: When \( \tan^{-1}(f(x)) \) is present, check if \( \frac{f'(x)}{1+(f(x))^2} \) (or a multiple of it) is also in the integrand, indicating a strong candidate for substitution.

Choose the Correct Option from the Given Alternatives for Questions 38 and 39:

Question 38. Integrate: \( \int \frac{10x^9 + 10^x \log_e 10}{x^{10} + 10^x} \, dx \)
(A) \( 10^x - x^{10} + C \)
(B) \( 10^x + x^{10} + C \)
(C) \( (10^x - x^{10})^{-1} + C \)
(D) \( \log (x^{10} + 10^x) + C \)
Answer: (D) \( \log (x^{10} + 10^x) + C \)
Let \( I = \int \frac{10x^9 + 10^x \log_e 10}{x^{10} + 10^x} \, dx \)
Let \( t = x^{10} + 10^x \).
Differentiating both sides with respect to \( x \):
Recall \( \frac{d}{dx}(x^n) = nx^{n-1} \) and \( \frac{d}{dx}(a^x) = a^x \log_e a \).
\( dt = (10x^{10-1} + 10^x \log_e 10) \, dx \)
\( dt = (10x^9 + 10^x \log_e 10) \, dx \).
Substitute these into the integral:
\( I = \int \frac{dt}{t} \)
The integral of \( \frac{1}{t} \) is \( \log |t| \).
\( I = \log |t| + c \)
Substitute back \( t = x^{10} + 10^x \):
\( I = \log |x^{10} + 10^x| + c \)
Therefore, option (D) is the correct choice.
In simple words: We solve this integral by substituting the entire denominator, \( x^{10} + 10^x \), with 't'. When we find the derivative of 't', it exactly matches the numerator. This simplifies the integral to a basic logarithmic form, which we solve and then substitute 't' back.

Exam Tip: If the numerator of a fraction is the exact derivative of the denominator, the integral will be the logarithm of the absolute value of the denominator. Remember the derivatives of \( x^n \) and \( a^x \).

Question 39. Integrate: \( \int \frac{d x}{\sin ^2 x \cos ^2 x} \)
(A) \( \tan x + \cot x + c \)
(B) \( \tan x - \cot x + c \)
(C) \( \tan x \cot x + c \)
(D) \( \tan x - \cot 2x + c \)
Answer: (B) \( \tan x - \cot x + c \)
Let \( I = \int \frac{1}{\sin^2 x \cos^2 x} \, dx \)
We know that \( \sin^2 x + \cos^2 x = 1 \). Substitute this into the numerator:
\( I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \, dx \)
Separate the terms:
\( I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) \, dx \)
Simplify each term:
\( I = \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) \, dx \)
Rewrite in terms of \( \sec^2 x \) and \( \csc^2 x \):
\( I = \int (\sec^2 x + \csc^2 x) \, dx \)
Integrate each term separately:
The integral of \( \sec^2 x \) is \( \tan x \).
The integral of \( \csc^2 x \) is \( -\cot x \).
\( I = \tan x - \cot x + c \)
Therefore, option (B) is the correct choice.
In simple words: To solve this, we replace the number '1' in the numerator with \( \sin^2 x + \cos^2 x \). Then we divide each term by the denominator, which simplifies to \( \sec^2 x + \csc^2 x \). We then integrate these standard trigonometric functions to get the final answer.

Exam Tip: For integrals involving \( \sin^2 x \cos^2 x \) in the denominator, a common trick is to replace the numerator with \( \sin^2 x + \cos^2 x \) to separate the fraction into easily integrable \( \sec^2 x \) and \( \csc^2 x \) terms.