Inter 2nd Year Maths Exercise 7d Solutions

Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 7 Integrals Solutions Exercise 7d will help students to clear their doubts quickly.

Inter 2nd Year Maths Integrals Solutions Exercise 7d

Integrals Exercise 7d Solutions

I. Integrate the functions

Question 1.
\(\frac{3 x^2}{x^6+1}\)
Solution:
Consider \(\frac{3 x^2}{x^6+1}\)
Let x³ = t ⇒ 3x² dx = dt
∴ ∫ \(\frac{3 x^2}{x^6+1}\) dx = ∫ \(\frac{\mathrm{dt}}{\mathrm{t}^2+1}\)
= tan^-1t + C = tan^-1 (x³) + C [∵ ∫ \(\frac{1}{x^2+a^2}\) dx = \(\frac{1}{a}\) tan^-1 \(\frac{x}{a}\) + C]

Question 2.
\(\frac{1}{\sqrt{1+4 x^2}}\)
Solution:
Consider \(\frac{1}{\sqrt{1+4 x^2}}\)
Let 2x = t ⇒ 2dx = dt

Question 3.
\(\frac{1}{\sqrt{(2-x)^2+1}}\)
Solution:

Question 4.
\(\frac{1}{\sqrt{9-25 x^2}}\)
Solution:

Question 5.
\(\frac{3 x}{1+2 x^4}\)
Solution:

Question 6.
\(\frac{x^2}{1-x^6}\)
Solution:
Consider \(\frac{x^2}{1-x^6}\)
Let x³ = t ⇒ 3x² dx = dt
∴ \(\int \frac{x^2}{1-x^6} d x=\frac{1}{3} \int \frac{d t}{1-t^2}\) [∵ \(\frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\) + C]
= \(\frac{1}{3}\left[\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|\right]\) + C
= \(\frac{1}{6} \log \left|\frac{1+x^3}{1-x^3}\right|\) + C

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

Question 8.
\(\frac{x^2}{\sqrt{x^6+a^6}}\)
Solution:
Consider \(\frac{x^2}{\sqrt{x^6+a^6}}\)
Let x³ = t ⇒ 3x² dx = dt
∴ \(\frac{x^2}{\sqrt{x^6+a^6}}\) dx = \(\frac{1}{3} \int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}^2+\left(\mathrm{a}^3\right)^2}}\)
= \(\frac{1}{3}\) log |t + \(\sqrt{t^2+\left(a^3\right)^2}\)| + C
[∵ \(\frac{\mathrm{dx}}{\sqrt{\mathrm{x}^2+\mathrm{a}^2}}=\log \left|\mathrm{x}+\sqrt{\mathrm{x}^2+\mathrm{a}^2}\right|\) + C]
= \(\frac{1}{3}\) log |x³ + \(\sqrt{x^6+a^6}\)| + C

Question 9.
\(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\)
Solution:
Consider \(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\)
Let tan x = t ⇒ sec² dx = dt
∴ \(\int \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\) dx
= latex]\int \frac{d t}{\sqrt{t^2+2^2}}[/latex]
= \(\log \left|t+\sqrt{t^2+4}\right|\) + C
[∵ \(\frac{\mathrm{dx}}{\sqrt{\mathrm{x}^2+\mathrm{a}^2}}=\log \left|\mathrm{x}+\sqrt{\mathrm{x}^2+\mathrm{a}^2}\right|\) + C]
= \(\log \left|\tan x+\sqrt{\tan ^2 x+4}\right|\) + C

Question 10.
\(\frac{\cos x}{\sqrt{4-\cos ^4 x}}\)
Solution:
Consider \(\frac{\cos x}{\sqrt{4-\cos ^4 x}}\)
= \(\frac{\cos x}{\sqrt{4-\left(\cos ^2 x\right)^2}}=\frac{\cos x}{\sqrt{4-\left(1-\sin ^2 x\right)^2}}\)
Let sin x = t ⇒ cos x dx
∴ I = \(\int \frac{\cos x}{\sqrt{4-\cos ^4 x}}\) dx
= \(\int \frac{d t}{\sqrt{2^2-(t)^2}}\)
Using standard result
I = \(\frac{1}{\sqrt{3}} \sin ^{-1}\left(\frac{t}{\sqrt{3}}\right)\) + C
Substitute t = sin x = \(\frac{1}{\sqrt{3}} \sin ^{-1}\left(\frac{t}{\sqrt{3}}\right)\) + C

II. Integrate the following functions

Question 1.
\(\frac{1}{\sqrt{x^2+2 x+2}}\)
Solution:

Question 2.
\(\frac{1}{9 x^2+6 x+5}\)
Solution:

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

Question 4.
\(\frac{1}{\sqrt{(x-1)(x-2)}}\)
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:

Question 6.
\(\frac{1}{\sqrt{(x-a)(x-b)}}\)
Solution:
Consider \(\frac{1}{\sqrt{(x-a)(x-b)}}\)
(x – a)(x – b) can be written as x² – (a + b)x + ab
∴ x² – (a + b)x + ab
= x² – (a + b)x + \(\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4}+ab\)
= \(\left[x-\left(\frac{a+b}{2}\right)\right]^2-\frac{(a-b)^2}{4}\)

Question 7.
\(\frac{x+2}{\sqrt{x^2-1}}\)
Solution:
∫ \(\frac{x+2}{\sqrt{x^2-1}}\) dx
= ∫ \(\frac{x}{\sqrt{x^2-1}}\) dx + ∫ \(\frac{2}{\sqrt{x^2-1}}\) dx ………….. (i)
Now, I₁ = ∫ \(\frac{x}{\sqrt{x^2-1}}\) dx
Let, x² – 1 = t ⇒ 2x dx = dt ⇒ dx = \(\frac{d t}{2 x}\)
I₁ = \(\int \frac{\mathrm{x}}{\sqrt{\mathrm{t}}} \times \frac{\mathrm{dt}}{2 \mathrm{x}}=\frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\frac{1}{2} \int \mathrm{t}^{-1 / 2} \mathrm{dt}\)
= \(\frac{1}{2}\) [2t^1/2] = \(\sqrt{t}\)
= \({\sqrt{x^2-1}}\) + C₁ [∵ t = x² – 1]
Now, I₂ = 2 ∫ \(\frac{1}{\sqrt{x^2-1}}\) dx = 2 log|x + \({\sqrt{x^2-1}}\)| + C₂
[∵ \(\int \frac{d x}{\sqrt{x^2-a^2}}=\sin ^{-1}\left(\frac{x}{a}\right)\) + C]
On Putting the values of I₁ & I₂ in eq (i) we get
I = \({\sqrt{x^2-1}}\) + 2 log|x + \({\sqrt{x^2-1}}\)| + C
Where, C = C₁ + C₂

Question 8.
\(\frac{4 x+1}{\sqrt{2 x^2+x-3}}\)
Solution:
Consider \(\frac{4 x+1}{\sqrt{2 x^2+x-3}}\)
Let 2x² + x – 3 = t
⇒ (4x + 1)dx = dt ⇒ dx = \(\frac{d t}{4 x+1}\)
∴ \(\int \frac{4 x+1}{\sqrt{2 x^2+x-3}}\) dx = \(\int \frac{4 x+1}{\sqrt{t}} \times \frac{d t}{4 x+1}\)
= \(\int \frac{1}{\sqrt{t}}\) dt
= 2\( \sqrt{t}\) + C = 2\(\sqrt{2 x^2+x-3}\) + C [∵ t = 2x² + x – 3]

III. Integrate the following functions

Question 1.
\(\frac{5 x-2}{1+2 x+3 x^2}\)
Solution:
∫\(\frac{5 x-2}{1+2 x+3 x^2}\) dx
Let 5x – 2 = A\(\frac{d}{d x}\)(1 + 2x + 3x²) + B
⇒ 5x – 2 = A(2 + 6x) + B ⇒ 5x – 2 = 6Ax + (2A + B)
Equating the coefficients of x and constant term on both sides, we get
5 = 6A ⇒ A = \(\frac{5}{6}\) and
2A + B = -2 ⇒ \(\frac{5}{3}\) + 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
I = \(\frac{5}{6}\)I₁ + (-\(\frac{11}{3}\))I₂ …………. (1)
Now, I₁ = \(\int \frac{2+6 x}{1+2 x+3 x^2}\)dx
Let 1 + 2x + 3x² = t ⇒ (2+ 6x)dx = dt
∴ I₁ = \(\int \frac{d t}{t}\) = log|t| + C₁
⇒ I₁ = log(1 + 2x + 3x²) + C₁ ……(2)
Also I₂ = \(\int \frac{1}{1+2 x+3 x^2} d x\)
1 + 2x + 3x² can be written as

Question 2.
\(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}\)
Solution:
∫ \(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}\) dx = ∫ \(\frac{6 x+7}{\sqrt{x^2-9 x+20}}\) dx
Let 6x + 7 = A\(\frac{d}{d x}\)(x² – 9x + 20) + B
⇒ 6x + 7 = A(2x – 9) + B
⇒ 6x + 7 = 2Ax + (-9A + B)
On equating the coefficients of x and constant term on both sides, we get
2A = 6 ⇒ A = 3; -9A + B = 7 ⇒ B = 34

Let x² – 9x + 20 = t ⇒ (2x – 9)dx = dt
∴ I₁ = \(\int \frac{d t}{\sqrt{t}}[latex] = 2[latex]\sqrt{t}\) + C₁ = 2\(\sqrt{x^2-9 x+20}\) + C₁ ……………..(2)
∴ I₂ = \(\int \frac{1}{\sqrt{x^2-9 x+20}}\) dx
x² – 9x + 20 can be written as
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 = \(\int \frac{1}{\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2}}\) dx
= log\(\left|\left(x-\frac{9}{2}\right)+\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{4}\right)}\right|\) + C₂
[∵ \(\int \frac{\mathrm{dx}}{\sqrt{\mathrm{x}^2-\mathrm{a}^2}}=\log \left|\mathrm{x}+\sqrt{\mathrm{x}^2-\mathrm{a}^2}\right|\)
= log \(\left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\) + C₂ ……………………… (3)
On Substituting the values of I₁ & I₂ for equations (2) and (3) in (1), we get
\(\int \frac{6 x+7}{\sqrt{x^2-9 x+20}}\) dx
= \(3\left[2 \sqrt{x^2-9 x+20}\right]+34 \log \left[\left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\right]\) + 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.
\(\frac{x+2}{\sqrt{4 x-x^2}}\)
Solution:
Let I = ∫ \(\frac{x+2}{\sqrt{4 x-x^2}}\) dx
Let x + 2 = A\(\frac{d}{d x}\)(4x – x²) + B ⇒ x + 2 = A(4 – 2x) + B
On 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
∴ I = \(\int \frac{x+2}{\sqrt{4 x-x^2}} d x=\int \frac{-\frac{1}{2}(4-2 x)+4}{\sqrt{\left(4 x-x^2\right)}} d x\)
= \(-\frac{1}{2} \int \frac{(4-2 x)}{\sqrt{\left(4 x-x^2\right)}} d x+4 \int \frac{1}{\sqrt{\left(4 x-x^2\right)}} d x\)
let I₁ = \(\int \frac{4-2 x}{\sqrt{4 x-x^2}} d x\) and I₂ = \(\int \frac{1}{\sqrt{4 x-x^2}} d x\)
∴ I = –\(\frac{1}{2}\)I₁ + 4I₂
Now, 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 = \(\int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=2 \sqrt{\mathrm{t}}\) + C₁ = \(2 \sqrt{4 \mathrm{x}-\mathrm{x}^2}\) + C₁ …….(2)
Now I₂ = \(\int \frac{1}{\sqrt{4 x-x^2}} d x\)
⇒ 4x – x² = -(-4x + x²)
= -{(x – 2)² – 4} = (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
\(\int \frac{x+2}{\sqrt{4 x-x^2}} d x=-\frac{1}{2}\left(2 \sqrt{4 x-x^2}\right)+4 \sin ^{-1}\left(\frac{x-2}{2}\right)\) + C₂ ……….. (3)
[∵ \(\int \frac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1}\left(\frac{x}{a}\right)\)]
On substituting the values of I₁ & I₂ from eq (2) & (3) in eq (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 [∵ –\(\frac{1}{2}\)C₁ + 4C₂ = C]

Question 4.
\(\frac{x+2}{\sqrt{x^2+2 x+3}}\)
Solution:
∫ \(\frac{x+2}{\sqrt{x^2+2 x+3}}\) dx
Let x + 2 = A\(\frac{d}{d x}\)(x² + 2x + 3) + B
⇒ x +2 = A(2x + 2) + B
⇒ x + 2 = 2Ax + (2A + B)
On equating the coefficient of x & constant term on both sides, we get
2A = 1 ⇒ A = \(\frac{1}{2}\) &
2A + B = 2 ⇒ 2 × \(\frac{1}{2}\) +B = 2 ⇒ B = 2 – 1 = 1
∴ (x + 2) = \(\frac{1}{2}\) (2x + 2) + 1

= log|x + 1 + \(\sqrt{(x+1)^2+2}\)| + C₂
= log|x + 1 + \(\sqrt{x^2+2 x+3}\)| + C₂ …………. (3)
On putting the values of I₁ & I₂ from eq(2) & (3) in eq (1), we get
∫ \(\frac{x+2}{\sqrt{x^2+2 x+3}}\) dx
= \(\frac{1}{2}\left[2 \sqrt{x^2+2 x+3}\right]+\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|\) + C
[∵ \(\frac{1}{2}\)C₁ + 4C₂ = C]
= \(\sqrt{x^2+2 x+3}+\log \mid(x+1)+\sqrt{x^2+2 x+3}\) + C

Question 5.
\(\frac{x+3}{x^2-2 x-5}\)
Solution:
∫ \(\frac{x+3}{x^2-2 x-5}\) dx
Let (x + 3) = A\(\frac{d}{d x}\)(x² – 2x – 5) + B
⇒ (x + 3) = A(2x – 2) + B
⇒ x + 3 = 2Ax – 2A + B
On 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

On substituting the values of I₂ & I₂ from eq(2) and (3) in (1), we get
∫ \(\frac{x+3}{x^2-2 x-5}\) dx
= \(\frac{1}{2} \log \left|x^2-2 x-5\right|+4\left[\frac{1}{2 \sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|\right]\)
[∵ \(\frac{1}{2}\)C₁ + 4C₂ = 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

Question 6.
\(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\)
Solution:
∫ \(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\) dx
Let 5x + 3 = A\(\frac{d}{d x}\)(x² + 4x + 10) + B
⇒ 5x + 3 = A(2x + 4) + B
⇒ 5x + 3 = 2Ax + 4A + B
On 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.
\(\frac{1}{x \sqrt{a x-x^2}}\) [Hint : Put x = \(\frac{a}{t}\)]
Solution:
Let I = ∫\(\frac{1}{x \sqrt{a x-x^2}}\) dx
Put x = \(\frac{a}{t}\) ⇒ dx = \(-\frac{a}{t^2}\)dt