AP Inter 2nd Year Maths Exercise 7d Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7d Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7d

I.

Question 1.
Find the integral of \(\frac{3 x^2}{x^6+1}\)
Solution:
Put x³ = t ⇒ 3x² dx = dt

Question 2.
Find the integral of \(\frac{3 x^2}{x^6+1}\)
Solution:
Put 2x = t ⇒ 2dx = dt

Question 3.
Find the integral of \(\frac{1}{\sqrt{(2-x)^2+1}}\)
Solution:
Put 2 – x = t ⇒ -dx = dt

Question 4.
Find the integral of \(\frac{1}{\sqrt{9-25 x^2}}\)
Solution:
Put 5x = t ⇒ 5dx = dt \(\left[ \int \frac{1}{\sqrt{a^2-x^2}} d x={Sin}^{-1}\left(\frac{x}{a}\right)+C\right]\)

Question 5.
Find the integral of \(\frac{3 x}{1+2 x^4}\)
Solution:
Put \(\sqrt{2}\)x² = t ⇒ 2\(\sqrt{2}\)xdx = dt \(\left[\int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C\right]\)

Question 6.
Find the integral of \(\frac{x^2}{1-x^6}\)
Solution:
Put x³ = t ⇒ 3x² dx = dt
∴ \(\int \frac{x^2}{1-x^6} d x=\frac{1}{3} \int \frac{d t}{1-t^2}=\frac{1}{3}\left[\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|\right]+C=\frac{1}{6} \log \left|\frac{1+x^3}{1-x^3}\right|+C\)

Question 7.
Find the integral of \(\frac{x-1}{\sqrt{x^2-1}}\)
Solution:

Question 8.
Find the integral of \(\frac{x^2}{\sqrt{x^6+a^6}}\)
Solution:
Put x³ = t ⇒ 3x² dx = dt \(\left[\int \frac{1}{\sqrt{x^2+a^2}} d x=\log \left|x+\sqrt{x^2+a^2}\right|\right]\)

Question 9.
Find the integral of \(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\)
Solution:
Put tan x = t ⇒ sec² dx = dt
∴ \(\int \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}} d x=\int \frac{d t}{\sqrt{t^2+2^2}}=\log \left|t+\sqrt{t^2+4}\right|+C=\log \left|\tan x+\sqrt{\tan ^2 x+4}\right|+C\)

Question 10.
Find the integral of \(\frac{\cos x}{\sqrt{4-\cos ^4 x}}\)
Solution:
Put sin x = t ⇒ cos xdx = dt
∴ \(\int \frac{\cos x}{\sqrt{4-\cos ^4 x}} d x=\int \frac{d t}{\sqrt{2^2-(t)^2}}=\sin ^{-1}\left(\frac{t}{2}\right)+C=\sin ^{-1}\left(\frac{\sin x}{2}\right)+C\)

II.

Question 1.
Find the integral of \(\frac{1}{\sqrt{x^2+2 x+2}}\)
Solution:
We have x² + 2x + 2 = x² + 2x + 1 – 1 + 2 = (x + 1)² + 1²

Question 2.
Find the integral of \(\frac{1}{9 x^2+6 x+5}\)
Solution:
We have 9x² + 6x + 5 = 9x² + 6x + 1 + 4 = (3x + 1)² + 2²

Question 3.
Find the integral of \(\frac{1}{\sqrt{7-6 x-x^2}}\)
Solution:
7 – 6x – x² can be written as 7 – (x² + 6x + 9 – 9) [∵ \(\int \frac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+C\)]
Thus 7 – (x² + 6x + 9 – 9) = 16 – (x² + 6x + 9) = 16 – (x + 3)² = 4² – (x + 3)²
∴ \(\int \frac{1}{\sqrt{7-6 x-x^2}} d x=\int \frac{1}{\sqrt{4^2-(x+3)^2}} d x\) Put x + 3 = t ⇒ dx = dt
∴ \(\int \frac{1}{\sqrt{4^2-(x+3)^2}} d x=\int \frac{1}{\sqrt{4^2-t^2}} d t=\sin ^{-1}\left(\frac{t}{4}\right)+C=\sin ^{-1}\left(\frac{x+3}{4}\right)+C\)

Question 4.
Find the integral of \(\frac{1}{\sqrt{(x-1)(x-2)}}\)
Solution:
We have (x – 1)(x – 1) = x² – 3x + 2

Question 5.
Find the integral of \(\frac{1}{\sqrt{8+3 x-x^2}}\)
Solution:
We have 8 + 3x – x² = -(x² – 3x – 8) = \(-\left(x^2-3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2-8\right)\)

Question 6.
Find the integral of \(\frac{1}{\sqrt{(x-a)(x-b)}}\)
Solution:
We have (x – a)(x – b) = x² – (a + b)x + ab
⇒ x² – (a + b)x + ab = x² – (a + b)x + \(\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4}+a b=\left[x-\left(\frac{a+b}{2}\right)\right]^2-\frac{(a-b)^2}{4}\)

Question 7.
Find the integral of \(\frac{x+2}{\sqrt{x^2-1}}\)
Solution:
Let x + 2 = A\(\frac{d}{d x}\)(x² – 1) + B …………..(1) ⇒ x + 2 = A(2x) + B
Equating the coefficientsof x and constant term on both sides, we get
2A = 1 ⇒ A = \(\frac{1}{2}\); B = 2
From (1) we get \(\int \frac{\mathrm{x}+2}{\sqrt{\mathrm{x}^2-1}} \mathrm{dx}=\int \frac{\frac{1}{2}(2 \mathrm{x})+2}{\sqrt{\mathrm{x}^2-1}} \mathrm{dx}=\frac{1}{2} \int \frac{2 \mathrm{x}}{\sqrt{\mathrm{x}^2-1}} \mathrm{dx}+\int \frac{2}{\sqrt{\mathrm{x}^2-1}} \mathrm{dx}\) ……………….(2)
To find \(\frac{1}{2} \int \frac{2 x}{\sqrt{x^2-1}} d x\) we take x² – 1 = t ⇒ 2xdx = dt

Question 8.
Find the integral of \(\frac{4 x+1}{\sqrt{2 x^2+x-3}}\)
Solution:
Put 2x² + x – 3 = t ⇒ (4x + 1)dx = dt
∴ \(\int \frac{4 x+1}{\sqrt{2 x^2+x-3}} d x=\int \frac{1}{\sqrt{t}} d t=2 \sqrt{t}+C=2 \sqrt{2 x^2+x-3}+C\)
\(\int \frac{f^{\prime}(x)}{\sqrt{f(x)}} d x=2 \sqrt{f(x)}+c\)

III.

Question 1.
Find the integral of \(\frac{5 x-2}{1+2 x+3 x^2}\)
Solution:
Let 5x – 2 = A\(\frac{d}{d x}\)(1 + 2x + 3x²) + B ⇒ 5x – 2 = A(2 + 6x) + B
Equating the coefficients of x and constant term on both sides, we get
5 = 6A ⇒ A = \(\frac{5}{6}\)
2A + B = -2 ⇒ B = \(-\frac{11}{3}\)
∴ 5x – 2 = \(\frac{5}{6}\)(2 + 6x) + (\(-\frac{11}{3}\))
∴ \(\int \frac{5 x-2}{1+2 x+3 x^2} d x=\int \frac{\frac{5}{6}(2+6 x)-\frac{11}{3}}{1+2 x+3 x^2} d x=\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^2} d x-\frac{11}{3} \int \frac{1}{1+2 x+3 x^2} d x\)
Let I₁ = \(\int \frac{2+6 x}{1+2 x+3 x^2}\)dx and I₂ = \(\int \frac{1}{1+2 x+3 x^2}\)dx
∴ \(\int \frac{5 x-2}{1+2 x+3 x^2} d x=\frac{5}{6} I_1-\frac{11}{3} I_2\) …..(1)
First we find I₁ = \(\int \frac{2+6 x}{1+2 x+3 x^2}\)dx
Put 1 + 2x + 3x² = t ⇒ (2+ 6x)dx = dt
∴ I₁ = \(\int \frac{d t}{t}\) ⇒ I₁ = log |t| ⇒ I₁ = log |1 + 2x + 3x² | ……(2)
Now I₂ = \(\int \frac{1}{1+2 x+3 x^2} d x\)
Here, 1 + 2x + 3x² = 1 + 3(x² + \(\frac{2}{3}\)x) [\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c]

Question 2.
Find the integral of \(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}\)
Solution:
We have \(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}=\frac{6 x+7}{\sqrt{x^2-9 x+20}}\)
Let 6x + 7 = A\(\frac{d}{d x}\)(x – 9x + 20) + B ⇒ 6x + 7 = A(2x – 9) + B
Equating the coefficients of x and constant term, we get
2A = 6 ⇒ A = 3; -9A + B = 7 ⇒ B = 34
∴ 6x+ 7 = 3(2x – 9) + 34

Let x² – 9x + 20 = t ⇒ (2x – 9)dx = dt
∴ I₁ = \(\int \frac{d t}{\sqrt{t}} \Rightarrow I_1=2 \sqrt{t} \Rightarrow I_1=2 \sqrt{x^2-9 x+20}\) ……………..(2)
Now I₂ = \(\int \frac{1}{\sqrt{x^2-9 x+20}}\)
x² – 9x + 20 = x² – 9x + 20 + \(\frac{81}{4}-\frac{81}{4}\)
x² – 9x + 20 + \(\frac{81}{4}-\frac{81}{4}=\left(x-\frac{9}{2}\right)^2-\frac{1}{4}=\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2\)
⇒ I₂ = \(\int \frac{1}{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2} d x=\log \left|\left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\right|\) ………….(3)
Substituting equations (2) and (3) in (1), we get
\(\int \frac{6 x+7}{\sqrt{x^2-9 x+20}} d x=3\left[2 \sqrt{x^2-9 x+20}\right]+34 \log \left[\left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\right]+C\)
= \(6 \sqrt{x^2-9 x+20}+34 \log \left[\left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\right]+C\)

Question 3.
Find the integral of \(\frac{x+2}{\sqrt{4 x-x^2}}\)
Solution:
Let x + 2 = A\(\frac{d}{d x}\)(4x – x²) + B
Equating the coefficients of x and constant term on both sides, we get
-2A = 1 ⇒ A = \(-\frac{1}{2}\); 4A + B = 2 ⇒ B = 4 ⇒ (x + 2) = \(-\frac{1}{2}\)(4 – 2x) + 4
∴ \(\int \frac{x+2}{\sqrt{4 x-x^2}} d x=\int \frac{-\frac{1}{2}(4-2 x)+4}{\sqrt{\left(4 x-x^2\right)}} d x=-\frac{1}{2} \int \frac{(4-2 x)}{\sqrt{\left(4 x-x^2\right)}} d x+4 \int \frac{1}{\sqrt{\left(4 x-x^2\right)}} d x\)
let I₁ = \(\int \frac{4-2 x}{\sqrt{4 x-x^2}} d x\) and I₂ = \(\int \frac{1}{\sqrt{4 x-x^2}} d x\)
∴ \(\int \frac{x+2}{\sqrt{4 x-x^2}} d x=-\frac{1}{2} I_1+4 I_2\) …….(1)
First we find I₁ = \(\int \frac{4-2 x}{\sqrt{4 x-x^2}} d x\) [∵ \(\int \frac{1}{\sqrt{x}} d x=2 \sqrt{x}+C\)]
Let 4x – x² = t ⇒ (4 – 2x)dx = dt ⇒ I₁ = \(\int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=2 \sqrt{\mathrm{t}}=2 \sqrt{4 \mathrm{x}-\mathrm{x}^2}\) …….(2)
Now I₂ = \(\int \frac{1}{\sqrt{4 x-x^2}} d x\)
⇒ 4x – x² = -(-4x + x²) = (-4x + x² + 4 – 4) = 4 – (x – 2)² = (2)² – (x – 2)²
∴ I₂ = \(\int \frac{1}{\sqrt{(2)^2-(x-2)^2}} d x=\sin ^{-1}\left(\frac{x-2}{2}\right)\) …..(3)
Substituting (2) and (3) in (1), we get
\(\int \frac{x+2}{\sqrt{4 x-x^2}} d x=-\frac{1}{2}\left(2 \sqrt{4 x-x^2}\right)+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+C\)
= \(-\sqrt{4 x-x^2}+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+C\)

Question 4.
Find the integral of \(\frac{x+2}{\sqrt{x^2+2 x+3}}\)
Solution:

First we find I₁ = \(\int \frac{2 x+2}{\sqrt{x^2+2 x+3}} d x\)
Put x² + 2x + 3 = t ⇒ (2x + 2)dx = dt
∴ I₁ = \(\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}=2 \sqrt{x^2+2 x+3}\) ………..(2)
Now I₂ = \(\int \frac{1}{\sqrt{x^2+2 x+3}} d x\),
Consider x² + 2x + 3 = x² + 2x + 1 + 2 = (x + 1)² + \((\sqrt{2})^2\)
Now I₂ = \(\int \frac{1}{\sqrt{(x+1)^2+(\sqrt{2})^2}} d x=\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|\) ….(3)
Substituting (2) and (3) in (1), we get
\(\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x=\frac{1}{2}\left[2 \sqrt{x^2+2 x+3}\right]+\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|+C\)
= \(\sqrt{x^2+2 x+3}+\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|+C\)

Question 5.
Find the integral of \(\frac{x+3}{x^2-2 x-5}\)
Solution:
Let (x + 3) = A\(\frac{d}{d x}\)(x² – 2x – 5) + B
⇒ (x + 3) = A(2x – 2) + B
Equating the coefficients of x and constant term on both sides, we get
2A = 1 ⇒ A = \(\frac{1}{2}\)
-2A + B = 3 ⇒ B = 4 ⇒ (x = 3) = \(\frac{1}{2}\)(2x – 2) + 4

Substituting (2) and (3) in (1), we get
\(\begin{aligned}
\int \frac{x+3}{x^2-2 x-5} d x & =\frac{1}{2} \log \left|x^2-2 x-5\right|+\frac{A}{\not 2 \sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+C \\
& =\frac{1}{2} \log \left|x^2-2 x-5\right|+\frac{2}{\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+C
\end{aligned}\)

Question 6.
Find the integral of \(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\)
Solution:
Let 5x + 3 = A\(\frac{d}{d x}\)(x² + 4x + 10) + B ⇒ 5x + 3 = A(2x + 4) + B
Equating the coefficients of x and constant term on both sides, we get 2A = 5 ⇒ A = \(\frac{5}{2}\)
4A + B = 3 ⇒ B = -7 ⇒ 5x + 3 = \(\frac{5}{2}\)(2x + 4) – 7

Question 7.
Find the integral of \(\frac{1}{x \sqrt{a x-x^2}}\) [Hint : Put x = \(\frac{a}{t}\)]
Solution:
Put x = \(\frac{a}{t}\) ⇒ dx = \(-\frac{a}{t^2}\)dt
∴ \(\int \frac{1}{x \sqrt{a x-x^2}} d x=\int \frac{1}{\frac{a}{t} \sqrt{a \cdot \frac{a}{t}-\left(\frac{a}{t}\right)^2}}\left(-\frac{a}{t^2} d t\right)=-\int \frac{1}{a t} \frac{1}{\sqrt{\frac{1}{t}-\frac{1}{t^2}}} d t=-\frac{1}{at} \int \frac{d t}{\frac{\sqrt{t-1}}{t}}\)
= \(-\frac{1}{a} \int \frac{1}{\sqrt{t-1}} d t=-\frac{1}{a}[2 \sqrt{t-1}]+C=-\frac{1}{a}\left[2 \sqrt{\frac{a}{x}-1}\right]+C=-\frac{2}{a}\left(\sqrt{\frac{a-x}{x}}\right)+C\)