RBSE Solutions Class 12 Maths Chapter 9 Integration Exercise 9.3

Integrate The Following Functions W.r.t. X.

Question 1. (a) \( \frac { 1 }{ 50+2x^2 } \)
(b) \( \frac { 1 }{ \sqrt{32-2x^2} } \)
Answer:
(a) We start by simplifying the denominator of the function:
\( \int \frac { 1 }{ 50+2x^2 } dx = \frac { 1 }{ 2 } \int \frac { 1 }{ 25+x^2 } dx \)
\( = \frac { 1 }{ 2 } \int \frac { 1 }{ x^2+5^2 } dx \)
Now, we use the standard integral formula \( \int \frac { 1 }{ x^2+a^2 } dx = \frac { 1 }{ a } \tan^{-1} \left( \frac { x }{ a } \right) + C \). Here, \( a=5 \).
\( = \frac { 1 }{ 2 } \cdot \frac { 1 }{ 5 } \tan^{-1} \left( \frac { x }{ 5 } \right) + C \)
\( = \frac { 1 }{ 10 } \tan^{-1} \left( \frac { x }{ 5 } \right) + C \)
(b) We simplify the expression under the square root in the denominator:
\( \int \frac { 1 }{ \sqrt{32-2x^2} } dx = \int \frac { 1 }{ \sqrt{2(16-x^2)} } dx \)
\( = \frac { 1 }{ \sqrt{2} } \int \frac { 1 }{ \sqrt{4^2-x^2} } dx \)
Now, we use the standard integral formula \( \int \frac { 1 }{ \sqrt{a^2-x^2} } dx = \sin^{-1} \left( \frac { x }{ a } \right) + C \). Here, \( a=4 \).
\( = \frac { 1 }{ \sqrt{2} } \sin^{-1} \left( \frac { x }{ 4 } \right) + C \)
In simple words: For part (a), we change the sum in the bottom of the fraction to match a known math pattern and then use a specific rule that turns it into an inverse tangent function. For part (b), we make the expression under the square root simpler to match another common pattern, then use a rule that gives an inverse sine function as the answer. Recognizing these standard integral forms is key to solving integration problems quickly.

🎯 Exam Tip: Always look for common factors or perfect squares in the denominator to simplify the integral into one of the standard forms. This makes solving much faster.

Question 2. (a) \( \frac { 1 }{ \sqrt{1-e^{2x}} } \)
(b) \( \frac { 1 }{ \sqrt{1+4x^2} } \)
Answer:
(a) To solve this integral, we use a substitution:
Let \( e^x = t \)
Differentiating both sides with respect to \( x \), we get \( e^x dx = dt \).
From this, we can express \( dx \) as \( dx = \frac{dt}{e^x} \). Since \( e^x = t \), then \( dx = \frac{dt}{t} \).
Now, substitute these into the integral:
\( \int \frac { 1 }{ \sqrt{1-e^{2x}} } dx = \int \frac { 1 }{ \sqrt{1-t^2} } \cdot \frac { dt }{ t } \)
\( = \int \frac { 1 }{ t\sqrt{1-t^2} } dt \)
Using the integral formula \( \int \frac{1}{x\sqrt{a^2-x^2}} dx = \frac{1}{a} \log \left| \frac{a - \sqrt{a^2-x^2}}{x} \right| \), where \( a=1 \), or a specific variation of it from the source:
\( = \log |1-\sqrt{1-t^2}| + C \)
Substitute \( t = e^x \) back into the expression:
\( = \log |1-\sqrt{1-e^{2x}}| + C \)
(b) We want to integrate \( \int \frac { 1 }{ \sqrt{1+4x^2} } dx \).
First, we rewrite \( 4x^2 \) as \( (2x)^2 \):
\( = \int \frac { 1 }{ \sqrt{1^2+(2x)^2} } dx \)
Now, we use a substitution:
Let \( u = 2x \). Differentiating both sides with respect to \( x \), we get \( du = 2 dx \), which means \( dx = \frac{du}{2} \).
Substitute these into the integral:
\( = \int \frac { 1 }{ \sqrt{1^2+u^2} } \frac{du}{2} \)
\( = \frac { 1 }{ 2 } \int \frac { 1 }{ \sqrt{1^2+u^2} } du \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{a^2+x^2} } dx = \log |x+\sqrt{a^2+x^2}| + C \). Here, \( a=1 \).
\( = \frac { 1 }{ 2 } \log |u+\sqrt{1^2+u^2}| + C \)
Substitute \( u = 2x \) back into the expression:
\( = \frac { 1 }{ 2 } \log |2x+\sqrt{1+4x^2}| + C \)
In simple words: For part (a), we replace \( e^x \) with a new letter to make the integral simpler. Then, we use a special rule that gives us a logarithm as the answer, and change it back to \( e^x \). For part (b), we make the bottom part of the fraction look like a common math pattern by using \( 2x \). After that, we use a simple swap and a known rule to get the logarithmic answer. Substitution is a powerful technique for converting complex integrals into simpler, recognizable forms.

🎯 Exam Tip: When dealing with square roots and exponential terms, look for substitutions that simplify the term inside the root or make the derivative appear in the numerator. Also, correctly applying logarithmic integral formulas is crucial.

Question 3. (a) \( \frac { 1 }{ \sqrt{a^2-b^2x^2} } \)
(b) \( \frac { 1 }{ \sqrt{(2-x)^2+1} } \)
Answer:
(a) We want to integrate \( \int \frac { 1 }{ \sqrt{a^2-b^2x^2} } dx \).
First, we use a substitution:
Let \( y = bx \). Differentiating both sides with respect to \( x \), we get \( dy = b dx \), which means \( dx = \frac{dy}{b} \).
Substitute these into the integral:
\( = \int \frac { 1 }{ \sqrt{a^2-y^2} } \frac{dy}{b} \)
\( = \frac { 1 }{ b } \int \frac { 1 }{ \sqrt{a^2-y^2} } dy \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{a^2-x^2} } dx = \sin^{-1} \left( \frac { x }{ a } \right) + C \).
\( = \frac { 1 }{ b } \sin^{-1} \left( \frac { y }{ a } \right) + C \)
Substitute \( y = bx \) back into the expression:
\( = \frac { 1 }{ b } \sin^{-1} \left( \frac { bx }{ a } \right) + C \)
(b) We want to integrate \( \int \frac { 1 }{ \sqrt{(2-x)^2+1} } dx \).
First, we use a substitution:
Let \( t = 2-x \). Differentiating both sides with respect to \( x \), we get \( dt = -dx \), which means \( dx = -dt \).
Substitute these into the integral:
\( = \int \frac { 1 }{ \sqrt{t^2+1^2} } (-dt) \)
\( = - \int \frac { 1 }{ \sqrt{t^2+1^2} } dt \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{x^2+a^2} } dx = \log |x+\sqrt{x^2+a^2}| + C \). Here, \( a=1 \).
\( = - \log |t+\sqrt{t^2+1^2}| + C \)
Substitute \( t = 2-x \) back into the expression:
\( = - \log |(2-x)+\sqrt{(2-x)^2+1}| + C \)
We can also expand the term under the square root:
\( (2-x)^2+1 = 4-4x+x^2+1 = x^2-4x+5 \)
So, the answer can also be written as:
\( = - \log |(2-x)+\sqrt{x^2-4x+5}| + C \)
In simple words: For part (a), we change \( bx \) to a new letter. This helps us use a common math rule that gives us an inverse sine as the answer. For part (b), we replace the \( (2-x) \) part with a new letter. This lets us use a standard rule for square root integrals to find the answer. We then put the \( (2-x) \) back in. Remember that the derivative of \( (c-x) \) is \( -1 \), which affects the sign of the resulting integral.

🎯 Exam Tip: When using substitution, always remember to differentiate the new variable to correctly replace \( dx \). Pay close attention to negative signs resulting from substitutions like \( u = c-x \).

Question 4. (a) \( \frac { x^2 }{ \sqrt{x^6+4} } \)
(b) \( \frac { x^4 }{ \sqrt{1-x^{10}} } \)
Answer:
(a) We want to integrate \( \int \frac { x^2 }{ \sqrt{x^6+4} } dx \).
First, we rewrite \( x^6 \) as \( (x^3)^2 \) and \( 4 \) as \( 2^2 \):
\( = \int \frac { x^2 }{ \sqrt{(x^3)^2+2^2} } dx \)
Now, we use a substitution:
Let \( t = x^3 \). Differentiating both sides with respect to \( x \), we get \( dt = 3x^2 dx \), which means \( x^2 dx = \frac{dt}{3} \).
Substitute these into the integral:
\( = \int \frac { 1 }{ \sqrt{t^2+2^2} } \frac{dt}{3} \)
\( = \frac { 1 }{ 3 } \int \frac { 1 }{ \sqrt{t^2+2^2} } dt \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{x^2+a^2} } dx = \log |x+\sqrt{x^2+a^2}| + C \). Here, \( a=2 \).
\( = \frac { 1 }{ 3 } \log |t+\sqrt{t^2+2^2}| + C \)
Substitute \( t = x^3 \) back into the expression:
\( = \frac { 1 }{ 3 } \log |x^3+\sqrt{x^6+4}| + C \)
(b) We want to integrate \( \int \frac { x^4 }{ \sqrt{1-x^{10}} } dx \).
First, we rewrite \( x^{10} \) as \( (x^5)^2 \) and \( 1 \) as \( 1^2 \):
\( = \int \frac { x^4 }{ \sqrt{1^2-(x^5)^2} } dx \)
Now, we use a substitution:
Let \( t = x^5 \). Differentiating both sides with respect to \( x \), we get \( dt = 5x^4 dx \), which means \( x^4 dx = \frac{dt}{5} \).
Substitute these into the integral:
\( = \int \frac { 1 }{ \sqrt{1^2-t^2} } \frac{dt}{5} \)
\( = \frac { 1 }{ 5 } \int \frac { 1 }{ \sqrt{1^2-t^2} } dt \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{a^2-x^2} } dx = \sin^{-1} \left( \frac { x }{ a } \right) + C \). Here, \( a=1 \).
\( = \frac { 1 }{ 5 } \sin^{-1} \left( \frac { t }{ 1 } \right) + C \)
Substitute \( t = x^5 \) back into the expression:
\( = \frac { 1 }{ 5 } \sin^{-1} (x^5) + C \)
In simple words: For part (a), we swap \( x^3 \) with a new letter, which makes the integral simpler. Then we use a specific math rule to solve it, giving us a logarithm as the answer. For part (b), we replace \( x^5 \) with a new letter to make the integral easy. This allows us to use a special rule that gives us an inverse sine as the answer. Look for powers of \( x \) where one is the derivative of another, as this often indicates a useful substitution.

🎯 Exam Tip: When the numerator contains a power of \( x \) that is one less than a power in the denominator (like \( x^2 \) and \( x^6 \), or \( x^4 \) and \( x^{10} \)), consider substituting the higher power term to simplify the integral.

Question 5. (a) \( \frac { 1 }{ x^2+6x+8 } \)
(b) \( \frac { 1 }{ \sqrt{2x^2-x+2} } \)
Answer:
(a) We want to integrate \( \int \frac { 1 }{ x^2+6x+8 } dx \).
First, we complete the square for the denominator \( x^2+6x+8 \):
\( x^2+6x+8 = (x^2+6x+9) - 9 + 8 = (x+3)^2 - 1 = (x+3)^2 - 1^2 \)
Substitute this back into the integral:
\( = \int \frac { 1 }{ (x+3)^2-1^2 } dx \)
Now, we use a substitution:
Let \( t = x+3 \). Differentiating both sides, we get \( dt = dx \).
Substitute these into the integral:
\( = \int \frac { 1 }{ t^2-1^2 } dt \)
We use the standard integral formula \( \int \frac { 1 }{ x^2-a^2 } dx = \frac { 1 }{ 2a } \log \left| \frac { x-a }{ x+a } \right| + C \). Here, \( a=1 \).
\( = \frac { 1 }{ 2 \cdot 1 } \log \left| \frac { t-1 }{ t+1 } \right| + C \)
Substitute \( t = x+3 \) back into the expression:
\( = \frac { 1 }{ 2 } \log \left| \frac { (x+3)-1 }{ (x+3)+1 } \right| + C \)
\( = \frac { 1 }{ 2 } \log \left| \frac { x+2 }{ x+4 } \right| + C \)
(b) We want to integrate \( \int \frac { 1 }{ \sqrt{2x^2-x+2} } dx \).
First, factor out \( \sqrt{2} \) from the denominator to make the coefficient of \( x^2 \) equal to 1:
\( = \frac { 1 }{ \sqrt{2} } \int \frac { 1 }{ \sqrt{x^2-\frac{1}{2}x+1} } dx \)
Now, complete the square for the expression under the square root, \( x^2-\frac{1}{2}x+1 \):
\( x^2-\frac{1}{2}x+1 = \left( x^2-\frac{1}{2}x+\left(\frac{1}{4}\right)^2 \right) - \left(\frac{1}{4}\right)^2 + 1 \)
\( = \left( x-\frac{1}{4} \right)^2 - \frac{1}{16} + 1 = \left( x-\frac{1}{4} \right)^2 + \frac{15}{16} = \left( x-\frac{1}{4} \right)^2 + \left( \frac{\sqrt{15}}{4} \right)^2 \)
Substitute this back into the integral:
\( = \frac { 1 }{ \sqrt{2} } \int \frac { 1 }{ \sqrt{\left( x-\frac{1}{4} \right)^2 + \left( \frac{\sqrt{15}}{4} \right)^2} } dx \)
Now, we use a substitution:
Let \( t = x-\frac{1}{4} \). Differentiating both sides, we get \( dt = dx \).
Substitute these into the integral:
\( = \frac { 1 }{ \sqrt{2} } \int \frac { 1 }{ \sqrt{t^2 + \left( \frac{\sqrt{15}}{4} \right)^2} } dt \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{x^2+a^2} } dx = \log |x+\sqrt{x^2+a^2}| + C \). Here, \( a = \frac{\sqrt{15}}{4} \).
\( = \frac { 1 }{ \sqrt{2} } \log \left| t + \sqrt{t^2 + \left( \frac{\sqrt{15}}{4} \right)^2} \right| + C \)
Substitute \( t = x-\frac{1}{4} \) back into the expression and simplify the term under the square root:
\( = \frac { 1 }{ \sqrt{2} } \log \left| \left( x-\frac{1}{4} \right) + \sqrt{x^2-\frac{1}{2}x+1} \right| + C \)
In simple words: For part (a), we change the bottom part of the fraction to a squared form by adding and subtracting a number. This helps us use a common rule that gives us a logarithm as the answer. For part (b), we first take out a number from the square root. Then, we change the inside part to a squared form. This helps us use a special rule that gives us a logarithm as the final answer. Completing the square is a fundamental technique for integrating rational functions and those with quadratic terms under a square root.

🎯 Exam Tip: When integrating functions with quadratic denominators or quadratic terms under a square root, completing the square is almost always the first and most important step to transform them into standard integral forms.

Question 6. (a) \( \frac { e^x }{ e^{2x}+2e^x \cos \alpha+1 } \)
(b) \( \frac { 1+\tan^2 x }{ \sqrt{\tan^2 x+3} } \)
Answer:
(a) We want to integrate \( \int \frac { e^x }{ e^{2x}+2e^x \cos \alpha+1 } dx \).
First, we rewrite the denominator by completing the square for \( e^x \):
\( e^{2x}+2e^x \cos \alpha+1 = (e^x)^2+2(e^x)(\cos \alpha) + \cos^2 \alpha - \cos^2 \alpha + 1 \)
\( = (e^x+\cos \alpha)^2 + (1-\cos^2 \alpha) = (e^x+\cos \alpha)^2 + \sin^2 \alpha \)
So the integral becomes:
\( = \int \frac { e^x }{ (e^x+\cos \alpha)^2 + \sin^2 \alpha } dx \)
Now, we use a substitution:
Let \( t = e^x+\cos \alpha \). Differentiating both sides with respect to \( x \), we get \( dt = e^x dx \) (since \( \cos \alpha \) is a constant).
Substitute these into the integral:
\( = \int \frac { dt }{ t^2+(\sin \alpha)^2 } \)
We use the standard integral formula \( \int \frac { 1 }{ x^2+a^2 } dx = \frac { 1 }{ a } \tan^{-1} \left( \frac { x }{ a } \right) + C \). Here, \( a = \sin \alpha \).
\( = \frac { 1 }{ \sin \alpha } \tan^{-1} \left( \frac { t }{ \sin \alpha } \right) + C \)
Substitute \( t = e^x+\cos \alpha \) back into the expression:
\( = \frac { 1 }{ \sin \alpha } \tan^{-1} \left( \frac { e^x+\cos \alpha }{ \sin \alpha } \right) + C \)
(b) We want to integrate \( \int \frac { 1+\tan^2 x }{ \sqrt{\tan^2 x+3} } dx \).
First, we use the trigonometric identity \( 1+\tan^2 x = \sec^2 x \):
\( = \int \frac { \sec^2 x }{ \sqrt{\tan^2 x+3} } dx \)
Now, we use a substitution:
Let \( t = \tan x \). Differentiating both sides with respect to \( x \), we get \( dt = \sec^2 x dx \).
Substitute these into the integral:
\( = \int \frac { dt }{ \sqrt{t^2+3} } = \int \frac { dt }{ \sqrt{t^2+(\sqrt{3})^2} } \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{x^2+a^2} } dx = \log |x+\sqrt{x^2+a^2}| + C \). Here, \( a=\sqrt{3} \).
\( = \log |t+\sqrt{t^2+(\sqrt{3})^2}| + C \)
Substitute \( t = \tan x \) back into the expression:
\( = \log |\tan x+\sqrt{\tan^2 x+3}| + C \)
In simple words: For part (a), we make the bottom part of the fraction into a squared sum. Then, we swap the complex part with a simple letter. This lets us use a rule that gives an inverse tangent as the answer. For part (b), we change \( 1+\tan^2 x \) to \( \sec^2 x \). Then, we replace \( \tan x \) with a new letter. This allows us to use a special rule for square root integrals to find the answer. Trigonometric identities are often essential tools for simplifying integrals involving trigonometric functions.

🎯 Exam Tip: For integrals involving \( e^x \) or trigonometric functions, look for ways to complete the square or use fundamental identities to simplify the integrand before applying substitution or standard formulas.

Question 7. (a) \( \frac { 1 }{ \sqrt{3x-2-x^2} } \)
(b) \( \frac { 1 }{ \sqrt{4+8x-5x^2} } \)
Answer:
(a) We want to integrate \( \int \frac { 1 }{ \sqrt{3x-2-x^2} } dx \).
First, we complete the square for the expression under the square root, \( 3x-2-x^2 \):
\( 3x-2-x^2 = -(x^2-3x+2) \)
To complete the square for \( x^2-3x+2 \):
\( x^2-3x+2 = \left( x^2-3x+\left(\frac{3}{2}\right)^2 \right) - \left(\frac{3}{2}\right)^2 + 2 \)
\( = \left( x-\frac{3}{2} \right)^2 - \frac{9}{4} + \frac{8}{4} = \left( x-\frac{3}{2} \right)^2 - \frac{1}{4} = \left( x-\frac{3}{2} \right)^2 - \left( \frac{1}{2} \right)^2 \)
Now, substitute this back into \( -(x^2-3x+2) \):
\( 3x-2-x^2 = - \left[ \left( x-\frac{3}{2} \right)^2 - \left( \frac{1}{2} \right)^2 \right] = \left( \frac{1}{2} \right)^2 - \left( x-\frac{3}{2} \right)^2 \)
So the integral becomes:
\( = \int \frac { 1 }{ \sqrt{\left( \frac{1}{2} \right)^2 - \left( x-\frac{3}{2} \right)^2} } dx \)
Now, we use a substitution:
Let \( t = x-\frac{3}{2} \). Differentiating both sides, we get \( dt = dx \).
Substitute these into the integral:
\( = \int \frac { 1 }{ \sqrt{\left( \frac{1}{2} \right)^2 - t^2} } dt \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{a^2-x^2} } dx = \sin^{-1} \left( \frac { x }{ a } \right) + C \). Here, \( a=\frac{1}{2} \).
\( = \sin^{-1} \left( \frac { t }{ \frac{1}{2} } \right) + C = \sin^{-1} (2t) + C \)
Substitute \( t = x-\frac{3}{2} \) back into the expression:
\( = \sin^{-1} \left( 2 \left( x-\frac{3}{2} \right) \right) + C = \sin^{-1} (2x-3) + C \)
(b) We want to integrate \( \int \frac { 1 }{ \sqrt{4+8x-5x^2} } dx \).
First, we complete the square for the expression under the square root, \( 4+8x-5x^2 \). We factor out \( -5 \) from the \( x \) terms:
\( 4+8x-5x^2 = -5 \left( x^2-\frac{8}{5}x-\frac{4}{5} \right) \)
To complete the square for \( x^2-\frac{8}{5}x-\frac{4}{5} \):
\( x^2-\frac{8}{5}x-\frac{4}{5} = \left( x^2-\frac{8}{5}x+\left(\frac{4}{5}\right)^2 \right) - \left(\frac{4}{5}\right)^2 - \frac{4}{5} \)
\( = \left( x-\frac{4}{5} \right)^2 - \frac{16}{25} - \frac{20}{25} = \left( x-\frac{4}{5} \right)^2 - \frac{36}{25} = \left( x-\frac{4}{5} \right)^2 - \left( \frac{6}{5} \right)^2 \)
Now, substitute this back into the expression \( -5 \left( x^2-\frac{8}{5}x-\frac{4}{5} \right) \):
\( 4+8x-5x^2 = -5 \left[ \left( x-\frac{4}{5} \right)^2 - \left( \frac{6}{5} \right)^2 \right] = 5 \left[ \left( \frac{6}{5} \right)^2 - \left( x-\frac{4}{5} \right)^2 \right] \)
So the integral becomes:
\( = \int \frac { 1 }{ \sqrt{5 \left[ \left( \frac{6}{5} \right)^2 - \left( x-\frac{4}{5} \right)^2 \right]} } dx \)
\( = \frac { 1 }{ \sqrt{5} } \int \frac { 1 }{ \sqrt{\left( \frac{6}{5} \right)^2 - \left( x-\frac{4}{5} \right)^2} } dx \)
Now, we use a substitution:
Let \( t = x-\frac{4}{5} \). Differentiating both sides, we get \( dt = dx \).
Substitute these into the integral:
\( = \frac { 1 }{ \sqrt{5} } \int \frac { 1 }{ \sqrt{\left( \frac{6}{5} \right)^2 - t^2} } dt \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{a^2-x^2} } dx = \sin^{-1} \left( \frac { x }{ a } \right) + C \). Here, \( a=\frac{6}{5} \).
\( = \frac { 1 }{ \sqrt{5} } \sin^{-1} \left( \frac { t }{ \frac{6}{5} } \right) + C \)
Substitute \( t = x-\frac{4}{5} \) back into the expression:
\( = \frac { 1 }{ \sqrt{5} } \sin^{-1} \left( \frac { x-\frac{4}{5} }{ \frac{6}{5} } \right) + C = \frac { 1 }{ \sqrt{5} } \sin^{-1} \left( \frac { 5x-4 }{ 6 } \right) + C \)
In simple words: For both parts, we change the inside of the square root into a simpler, squared form. This often involves carefully handling minus signs and factoring out constants. Once it matches the pattern \( a^2-t^2 \), we use a special rule that gives us an inverse sine as the answer. Be extra careful with negative signs when completing the square, especially when \( x^2 \) has a negative coefficient.

🎯 Exam Tip: For integrals involving quadratic expressions under a square root with a negative \( x^2 \) term, always factor out the negative sign (or the coefficient of \( x^2 \)) before completing the square to convert it into the \( a^2-x^2 \) form.

Question 8. (a) \( \frac { \sin x + \cos x }{ \sqrt{\sin 2x} } \)
(b) \( \frac { 1 }{ \sqrt{x^2+2ax+b^2} } \)
Answer:
(a) We want to integrate \( \int \frac { \sin x + \cos x }{ \sqrt{\sin 2x} } dx \).
First, we use a substitution that simplifies both the numerator and the term under the square root:
Let \( t = \sin x - \cos x \).
Differentiating both sides with respect to \( x \), we get \( dt = (\cos x - (-\sin x)) dx = (\cos x + \sin x) dx \). This matches the numerator.
To simplify the term under the square root, \( \sin 2x \), we square \( t \):
\( t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x \)
Since \( \sin^2 x + \cos^2 x = 1 \) and \( 2 \sin x \cos x = \sin 2x \):
\( t^2 = 1 - \sin 2x \)
So, \( \sin 2x = 1 - t^2 \).
Substitute these into the integral:
\( = \int \frac { dt }{ \sqrt{1-t^2} } \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{a^2-x^2} } dx = \sin^{-1} \left( \frac { x }{ a } \right) + C \). Here, \( a=1 \).
\( = \sin^{-1} (t) + C \)
Substitute \( t = \sin x - \cos x \) back into the expression:
\( = \sin^{-1} (\sin x - \cos x) + C \)
(b) We want to integrate \( \int \frac { 1 }{ \sqrt{x^2+2ax+b^2} } dx \).
First, we complete the square for the expression under the square root, \( x^2+2ax+b^2 \):
\( x^2+2ax+b^2 = (x^2+2ax+a^2) - a^2 + b^2 = (x+a)^2 + (b^2-a^2) \)
So the integral becomes:
\( = \int \frac { 1 }{ \sqrt{(x+a)^2 + (b^2-a^2)} } dx \)
Now, we use a substitution:
Let \( t = x+a \). Differentiating both sides, we get \( dt = dx \).
Substitute these into the integral:
\( = \int \frac { 1 }{ \sqrt{t^2 + (b^2-a^2)} } dt \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{x^2 \pm k^2} } dx = \log |x+\sqrt{x^2 \pm k^2}| + C \). Here, \( k^2 = b^2-a^2 \).
\( = \log |t+\sqrt{t^2+(b^2-a^2)}| + C \)
Substitute \( t = x+a \) back into the expression:
\( = \log |(x+a)+\sqrt{(x+a)^2+(b^2-a^2)}| + C \)
Since \( (x+a)^2+(b^2-a^2) \) simplifies back to \( x^2+2ax+b^2 \):
\( = \log |(x+a)+\sqrt{x^2+2ax+b^2}| + C \)
In simple words: For part (a), we swap \( (\sin x - \cos x) \) for a new letter. This changes the integral into a simple form that we solve using the inverse sine function. For part (b), we change the inside of the square root to a squared sum. This allows us to use a special rule that gives us a logarithm as the answer. For integrals involving \( \sin x + \cos x \) and \( \sin 2x \), the substitution \( t = \sin x \pm \cos x \) is very common and effective.

🎯 Exam Tip: When faced with trigonometric integrals, sometimes a substitution that simplifies to a simple polynomial form is more effective than direct trigonometric identities. Completing the square is crucial for quadratic expressions in denominators or under square roots.

Question 9. (a) \( \sqrt{\frac{a-x}{x}} \)
(b) \( \sqrt{\frac{a+x}{a-x}} \)
Answer:
(a) We want to integrate \( I = \int \sqrt{\frac{a-x}{x}} dx \).
For this type of integral, a trigonometric substitution is often effective:
Let \( x = a \sin^2 \theta \).
Differentiating both sides with respect to \( \theta \), we get \( dx = a \cdot 2 \sin \theta \cos \theta d\theta = 2a \sin \theta \cos \theta d\theta \).
Now, substitute these into the integral:
\( I = \int \sqrt{\frac{a-a \sin^2 \theta}{a \sin^2 \theta}} (2a \sin \theta \cos \theta) d\theta \)
Simplify the term inside the square root:
\( = \int \sqrt{\frac{a(1-\sin^2 \theta)}{a \sin^2 \theta}} (2a \sin \theta \cos \theta) d\theta = \int \sqrt{\frac{\cos^2 \theta}{\sin^2 \theta}} (2a \sin \theta \cos \theta) d\theta \)
\( = \int \frac{\cos \theta}{\sin \theta} (2a \sin \theta \cos \theta) d\theta = \int 2a \cos^2 \theta d\theta \)
Use the trigonometric identity \( \cos^2 \theta = \frac{1+\cos 2\theta}{2} \):
\( = \int 2a \left( \frac{1+\cos 2\theta}{2} \right) d\theta = a \int (1+\cos 2\theta) d\theta \)
Integrate term by term:
\( = a \left[ \theta + \frac{\sin 2\theta}{2} \right] + C \)
\( = a \theta + a \sin \theta \cos \theta + C \)
Finally, convert the expression back to terms of \( x \):
From \( x = a \sin^2 \theta \), we have \( \sin^2 \theta = \frac{x}{a} \). So, \( \sin \theta = \sqrt{\frac{x}{a}} \).
From \( \sin \theta = \sqrt{\frac{x}{a}} \), we can find \( \theta = \sin^{-1} \left( \sqrt{\frac{x}{a}} \right) \).
Also, \( \cos \theta = \sqrt{1-\sin^2 \theta} = \sqrt{1-\frac{x}{a}} = \sqrt{\frac{a-x}{a}} \).
Substitute these back into the integrated expression:
\( I = a \sin^{-1} \left( \sqrt{\frac{x}{a}} \right) + a \left( \sqrt{\frac{x}{a}} \right) \left( \sqrt{\frac{a-x}{a}} \right) + C \)
\( = a \sin^{-1} \left( \sqrt{\frac{x}{a}} \right) + \sqrt{x(a-x)} + C \)
(b) We want to integrate \( I = \int \sqrt{\frac{a+x}{a-x}} dx \).
For this type of integral, a trigonometric substitution is effective:
Let \( x = a \cos \theta \).
Differentiating both sides with respect to \( \theta \), we get \( dx = -a \sin \theta d\theta \).
Now, substitute these into the integral:
\( I = \int \sqrt{\frac{a+a \cos \theta}{a-a \cos \theta}} (-a \sin \theta) d\theta \)
Simplify the term inside the square root:
\( = \int \sqrt{\frac{a(1+\cos \theta)}{a(1-\cos \theta)}} (-a \sin \theta) d\theta \)
Use the half-angle identities \( 1+\cos \theta = 2\cos^2 \frac{\theta}{2} \) and \( 1-\cos \theta = 2\sin^2 \frac{\theta}{2} \):
\( = \int \sqrt{\frac{2\cos^2 \frac{\theta}{2}}{2\sin^2 \frac{\theta}{2}}} (-a \sin \theta) d\theta = \int \sqrt{\cot^2 \frac{\theta}{2}} (-a \sin \theta) d\theta \)
\( = \int \cot \frac{\theta}{2} (-a \sin \theta) d\theta \)
Use the identity \( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \):
\( = \int \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} (-a \cdot 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}) d\theta \)
\( = \int -2a \cos^2 \frac{\theta}{2} d\theta \)
Use the identity \( \cos^2 \frac{\theta}{2} = \frac{1+\cos \theta}{2} \):
\( = \int -2a \left( \frac{1+\cos \theta}{2} \right) d\theta = -a \int (1+\cos \theta) d\theta \)
Integrate term by term:
\( = -a [\theta + \sin \theta] + C \)
Finally, convert the expression back to terms of \( x \):
From \( x = a \cos \theta \), we have \( \cos \theta = \frac{x}{a} \). So, \( \theta = \cos^{-1} \left( \frac{x}{a} \right) \).
Also, \( \sin \theta = \sqrt{1-\cos^2 \theta} = \sqrt{1-\frac{x^2}{a^2}} = \frac{\sqrt{a^2-x^2}}{a} \).
Substitute these back into the integrated expression:
\( I = -a \cos^{-1} \left( \frac{x}{a} \right) - a \left( \frac{\sqrt{a^2-x^2}}{a} \right) + C \)
\( = -a \cos^{-1} \left( \frac{x}{a} \right) - \sqrt{a^2-x^2} + C \)
In simple words: For both parts, we use trigonometric substitutions to simplify the integral. For part (a), replacing \( x \) with \( a \sin^2 \theta \) helps. For part (b), replacing \( x \) with \( a \cos \theta \) simplifies it. These substitutions convert the complex integrals into simpler trigonometric forms, which are then integrated. Finally, the answer is converted back to terms of \( x \). Trigonometric substitutions are very powerful for integrals involving expressions like \( \sqrt{a-x} \) or \( \sqrt{a^2-x^2} \).

🎯 Exam Tip: For integrals of the form \( \sqrt{\frac{a-x}{x}} \) or \( \sqrt{\frac{a+x}{a-x}} \), trigonometric substitutions like \( x=a \sin^2 \theta \) or \( x=a \cos \theta \) are often the most effective. Remember to use double-angle or half-angle identities for simplification.

Question 10. (a) \( \frac { \sqrt{x} }{ \sqrt{a^3-x^3} } \)
(b) \( \frac { 1 }{ (a^2+x^2)^{3/2} } \)
Answer:
(a) We want to integrate \( I = \int \frac { \sqrt{x} }{ \sqrt{a^3-x^3} } dx \).
First, we rewrite the terms under the square root to identify a suitable substitution:
We can express \( x^3 \) as \( (x^{3/2})^2 \) and \( a^3 \) as \( (a^{3/2})^2 \).
So the integral is \( \int \frac { \sqrt{x} }{ \sqrt{(a^{3/2})^2-(x^{3/2})^2} } dx \)
Now, we use a substitution:
Let \( t = x^{3/2} \).
Differentiating both sides with respect to \( x \), we get \( dt = \frac{3}{2} x^{3/2-1} dx = \frac{3}{2} x^{1/2} dx = \frac{3}{2} \sqrt{x} dx \).
From this, we can write \( \sqrt{x} dx = \frac{2}{3} dt \).
Substitute these into the integral:
\( I = \int \frac { 1 }{ \sqrt{(a^{3/2})^2-t^2} } \left( \frac{2}{3} dt \right) \)
\( = \frac{2}{3} \int \frac { 1 }{ \sqrt{(a^{3/2})^2-t^2} } dt \)
We use the standard integral formula \( \int \frac { 1 }{ \sqrt{k^2-x^2} } dx = \sin^{-1} \left( \frac { x }{ k } \right) + C \). Here, \( k=a^{3/2} \).
\( = \frac{2}{3} \sin^{-1} \left( \frac{t}{a^{3/2}} \right) + C \)
Substitute \( t = x^{3/2} \) back into the expression:
\( = \frac{2}{3} \sin^{-1} \left( \frac{x^{3/2}}{a^{3/2}} \right) + C = \frac{2}{3} \sin^{-1} \left( \left( \frac{x}{a} \right)^{3/2} \right) + C \)
(b) We want to integrate \( I = \int \frac { 1 }{ (a^2+x^2)^{3/2} } dx \).
For this type of integral, a trigonometric substitution is effective:
Let \( x = a \tan \theta \).
Differentiating both sides with respect to \( \theta \), we get \( dx = a \sec^2 \theta d\theta \).
Substitute these into the integral:
\( I = \int \frac { a \sec^2 \theta }{ (a^2+(a \tan \theta)^2)^{3/2} } d\theta \)
Simplify the denominator:
\( = \int \frac { a \sec^2 \theta }{ (a^2+a^2 \tan^2 \theta)^{3/2} } d\theta = \int \frac { a \sec^2 \theta }{ (a^2(1+\tan^2 \theta))^{3/2} } d\theta \)
Using the identity \( 1+\tan^2 \theta = \sec^2 \theta \):
\( = \int \frac { a \sec^2 \theta }{ (a^2 \sec^2 \theta)^{3/2} } d\theta = \int \frac { a \sec^2 \theta }{ (a^2)^{3/2} (\sec^2 \theta)^{3/2} } d\theta \)
\( = \int \frac { a \sec^2 \theta }{ a^3 \sec^3 \theta } d\theta = \int \frac { 1 }{ a^2 \sec \theta } d\theta \)
\( = \frac{1}{a^2} \int \frac{1}{\sec \theta} d\theta = \frac{1}{a^2} \int \cos \theta d\theta \)
Integrate \( \cos \theta \):
\( = \frac{1}{a^2} \sin \theta + C \)
Finally, convert the expression back to terms of \( x \):
From \( x = a \tan \theta \), we have \( \tan \theta = \frac{x}{a} \).
Draw a right triangle with opposite side \( x \) and adjacent side \( a \). The hypotenuse will be \( \sqrt{x^2+a^2} \).
So, \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2+a^2}} \).
Substitute this back into the expression:
\( I = \frac{1}{a^2} \frac{x}{\sqrt{x^2+a^2}} + C \)
In simple words: For part (a), we replace \( x^{3/2} \) with a new letter. This makes the integral simpler and allows us to use a rule that gives an inverse sine as the answer. For part (b), we replace \( x \) with \( a \tan \theta \). This makes the integral much simpler to solve, giving us a final answer that includes \( x \) and a square root. Look for relationships between powers of \( x \), such as \( \sqrt{x} \) being the derivative of \( x^{3/2} \), to identify suitable substitutions.

🎯 Exam Tip: For integrals involving \( (a^n \pm x^n) \) or \( (a^2+x^2)^{n/2} \), look for substitutions like \( t=x^{n/2} \) or trigonometric substitutions like \( x=a \tan \theta \) (for \( a^2+x^2 \)) to simplify them into standard forms.

Question 11. (a) \( \frac { 1 }{ (1-x^2)^{3/2} } \)
(b) \( \frac { x+1 }{ \sqrt{x^2+1} } \)
Answer:
(a) We want to integrate \( I = \int \frac { 1 }{ (1-x^2)^{3/2} } dx \).
For this type of integral, a trigonometric substitution is often effective:
Let \( x = \sin \theta \).
Differentiating both sides with respect to \( \theta \), we get \( dx = \cos \theta d\theta \).
Now, substitute these into the integral:
\( I = \int \frac { \cos \theta }{ (1-\sin^2 \theta)^{3/2} } d\theta \)
Using the identity \( 1-\sin^2 \theta = \cos^2 \theta \):
\( = \int \frac { \cos \theta }{ (\cos^2 \theta)^{3/2} } d\theta = \int \frac { \cos \theta }{ \cos^3 \theta } d\theta \)
\( = \int \frac { 1 }{ \cos^2 \theta } d\theta = \int \sec^2 \theta d\theta \)
Integrate \( \sec^2 \theta \):
\( = \tan \theta + C \)
Finally, convert the expression back to terms of \( x \):
From \( x = \sin \theta \), we can draw a right triangle. The opposite side is \( x \), and the hypotenuse is \( 1 \). The adjacent side will be \( \sqrt{1^2-x^2} = \sqrt{1-x^2} \).
So, \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{\sqrt{1-x^2}} \).
Substitute this back into the integrated expression:
\( I = \frac{x}{\sqrt{1-x^2}} + C \)
(b) We want to integrate \( I = \int \frac { x+1 }{ \sqrt{x^2+1} } dx \).
We can split this integral into two simpler integrals:
\( I = \int \frac { x }{ \sqrt{x^2+1} } dx + \int \frac { 1 }{ \sqrt{x^2+1} } dx \)
Let's evaluate each integral separately.
For the first integral, \( I_1 = \int \frac { x }{ \sqrt{x^2+1} } dx \):
Let \( t = x^2+1 \). Differentiating both sides, we get \( dt = 2x dx \), which means \( x dx = \frac{dt}{2} \).
Substitute these into \( I_1 \):
\( I_1 = \int \frac { 1 }{ \sqrt{t} } \frac{dt}{2} = \frac{1}{2} \int t^{-1/2} dt \)
Integrate \( t^{-1/2} \):
\( = \frac{1}{2} \frac{t^{-1/2+1}}{-1/2+1} + C_1 = \frac{1}{2} \frac{t^{1/2}}{1/2} + C_1 = t^{1/2} + C_1 = \sqrt{t} + C_1 \)
Substitute \( t = x^2+1 \) back:
\( I_1 = \sqrt{x^2+1} + C_1 \)
For the second integral, \( I_2 = \int \frac { 1 }{ \sqrt{x^2+1} } dx \):
This is a standard integral of the form \( \int \frac { 1 }{ \sqrt{x^2+a^2} } dx = \log |x+\sqrt{x^2+a^2}| + C \). Here, \( a=1 \).
\( I_2 = \log |x+\sqrt{x^2+1^2}| + C_2 = \log |x+\sqrt{x^2+1}| + C_2 \)
Now, combine \( I_1 \) and \( I_2 \) to get the total integral \( I \):
\( I = I_1 + I_2 = (\sqrt{x^2+1} + C_1) + (\log |x+\sqrt{x^2+1}| + C_2) \)
\( = \sqrt{x^2+1} + \log |x+\sqrt{x^2+1}| + C \), where \( C = C_1+C_2 \).
In simple words: For part (a), we replace \( x \) with a sine function. This makes the integral very simple to solve using trigonometric identities. Then, we change the answer back to using \( x \). For part (b), we break the problem into two easier parts. We solve the first part by swapping a letter, and the second part using a known rule for square roots. Then we add the two answers together. Always consider splitting complex fractions or sums into multiple, simpler integrals if it helps with easier evaluation.

🎯 Exam Tip: When an integral has a sum in the numerator, splitting it into separate integrals for each term can often simplify the problem significantly, allowing the use of different integration techniques for each part.

Question 12. Integrate the following functions w.r.t. x.
(a) \( \frac { 1 }{ \sqrt { (x-\alpha)(\beta-x) } } \)
(b) \( \frac { 1 }{ \sqrt { 2x-x^2 } } \)
Answer:
(a) For the integral \( \int \frac { 1 }{ \sqrt { (x-\alpha)(\beta-x) } } dx \):
We use the substitution \( x = \alpha \cos^2 \theta + \beta \sin^2 \theta \).
This means \( dx = (\beta - \alpha) \sin 2\theta d\theta \).
Now, let's find \( x-\alpha \) and \( \beta-x \):
\( x-\alpha = (\alpha \cos^2 \theta + \beta \sin^2 \theta) - \alpha = \beta \sin^2 \theta - \alpha \sin^2 \theta = (\beta-\alpha) \sin^2 \theta \)
\( \beta-x = \beta - (\alpha \cos^2 \theta + \beta \sin^2 \theta) = \beta \cos^2 \theta - \alpha \cos^2 \theta = (\beta-\alpha) \cos^2 \theta \)
So, \( \sqrt{(x-\alpha)(\beta-x)} = \sqrt{(\beta-\alpha)^2 \sin^2 \theta \cos^2 \theta} = (\beta-\alpha) \sin \theta \cos \theta \).
Substituting these into the integral:
\( \int \frac { 1 }{ (\beta-\alpha) \sin \theta \cos \theta } (\beta-\alpha) \sin 2\theta d\theta \)
We know that \( \sin 2\theta = 2 \sin \theta \cos \theta \).
\( \implies \int \frac { (\beta-\alpha) (2 \sin \theta \cos \theta) }{ (\beta-\alpha) \sin \theta \cos \theta } d\theta \)
\( \implies \int 2 d\theta \)
This simplifies to \( 2\theta + C \).
From our substitution, \( x = \alpha \cos^2 \theta + \beta \sin^2 \theta \), we can also write \( x-\alpha = (\beta-\alpha) \sin^2 \theta \).

\( \implies \sin^2 \theta = \frac{x-\alpha}{\beta-\alpha} \)

\( \implies \sin \theta = \sqrt{\frac{x-\alpha}{\beta-\alpha}} \)

\( \implies \theta = \sin^{-1} \sqrt{\frac{x-\alpha}{\beta-\alpha}} \)
Therefore, the final result is \( 2 \sin^{-1} \sqrt{\frac{x-\alpha}{\beta-\alpha}} + C \). This formula is very useful for integrals involving square roots of quadratic expressions.
(b) For the integral \( \int \frac { 1 }{ \sqrt { 2x-x^2 } } dx \):
First, we complete the square for the term inside the square root, \( 2x-x^2 \).
We can write \( 2x-x^2 = -(x^2-2x) \).
To complete the square for \( x^2-2x \), we add and subtract \( (2/2)^2 = 1^2 = 1 \).
So, \( x^2-2x = (x^2-2x+1) - 1 = (x-1)^2 - 1 \).
Now substitute this back into the original expression:
\( 2x-x^2 = -((x-1)^2 - 1) = 1 - (x-1)^2 \).
The integral becomes \( \int \frac { 1 }{ \sqrt { 1 - (x-1)^2 } } dx \).
This is in the standard form \( \int \frac { 1 }{ \sqrt { a^2-u^2 } } du = \sin^{-1} \left( \frac{u}{a} \right) + C \).
Here, \( a=1 \) and \( u=(x-1) \).
So, the solution is \( \sin^{-1} (x-1) + C \). This technique of completing the square is key to solving many integrals involving quadratic expressions.
In simple words: For part (a), we use a special trick called substitution to change the integral into a simpler form, then solve it and change it back. For part (b), we rearrange the bottom part of the fraction to make it fit a known integration formula, which then gives us the answer directly.

🎯 Exam Tip: For integrals of the form \( \frac { 1 }{ \sqrt { (x-a)(b-x) } } \), remember the standard result \( 2 \sin^{-1} \sqrt{\frac{x-a}{b-a}} + C \). For integrals like \( \frac { 1 }{ \sqrt { Ax^2+Bx+C } } \), always complete the square in the denominator to simplify it into a standard form like \( \sqrt{a^2-u^2} \), \( \sqrt{u^2-a^2} \), or \( \sqrt{u^2+a^2} \).