Inter 2nd Year Maths Exercise 7h Solutions

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

Inter 2nd Year Maths Integrals Solutions Exercise 7h

Integrals Exercise 7h Solutions

I. Evaluate the following definite integrals

Question 1.
\(\int_{-1}^1(x+1) d x\)
Solution:
Let I = \(\int_{-1}^1(x+1) d x=\left[\frac{x^2}{2}+x\right]_{-1}^1\)
= \(\left[\frac{1}{2}+1\right]\) – \(\left(\frac{(-1)^2}{2}+(-1)\right)\)
= \(\frac{3}{2}+\frac{1}{2}=\frac{4}{2}\) = 2

Question 2.
Evaluate \(\int_2^3 \frac{1}{x} d x\)
Solution:
Let I = \(\int_2^3 \frac{1}{x}\) dx
= \((\log \mathrm{x})_2^3\)
= log 3 – log 2 = log \(\frac{3}{2}\)

Question 3.
\(\int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x\)
Solution:
Let I = \(\int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x\)
\(\left(\frac{4 x^4}{4}-\frac{5 x^3}{3}+\frac{6 x^2}{2}+9 x\right)_1^2=\left(x^4-\frac{5 x^3}{3}+3 x^2+9 x\right)_1^2\)
= \(\left(2^4-\frac{5(2)^3}{3}+3(2)^2+9(2)\right)-\left(1-\frac{5}{3}+3+9\right)\)
= \(16-\frac{40}{3}+12+18-\left(13-\frac{5}{3}\right)=\left(46-\frac{40}{3}\right)-\left(13-\frac{5}{3}\right)\)
= \(\frac{138-40}{3}-\left(\frac{39-5}{3}\right)=\frac{98}{3}-\frac{34}{3}=\frac{64}{3}\).

Question 4.
\(\int_0^\pi 4 \sin 2 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{4}} \sin 2 x d x=\left(\frac{-\cos 2 x}{2}\right)_0^{\pi / 4}\)
= \(\frac{-\cos 2 \frac{\pi}{4}}{2}+\frac{\cos (0)}{2}\)
= 0 +\(\frac{1}{2}\) = \(\frac{1}{2}\)

Question 5.
\(\int_0^\pi 2 \cos 2 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \cos 2 x d x=\left[\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}}\)
= \(\frac{1}{2}\) (sin π – sin 0) = 0

Question 6.
\(\int_4^5 e^x d x\)
Solution:
Let I = \(\int_4^5 e^x d x=\left[e^x\right]_4^5\)
= e⁵ – e⁴ = e⁴(e – 1)

Question 7.
\(\int_0^\pi 4 \tan x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{4}} \tan x d x=[-\log \cos x]_0^{\pi / 4}\)
= – log|cos \( \frac{\pi}{4}\) |+ log |cos 0|
= – log \( \frac{1}{\sqrt{2}}\) + log 1
= – log 2^1/2 + 0 = \( \frac{1}{2}\) log 2

Question 8.
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} {cosec} x d x\)
Solution:
Let I = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} {cosec} x d x=[\log |{cosec} x-\cot x|]_{\pi / 6}^{\pi / 4}\)
= log |cosec \( \frac{\pi}{4}\) – cot \(\frac{\pi}{4}\)| – [log|cosec \(\frac{\pi}{6}\) – cot\(\frac{\pi}{6}\)|]
= \(\log |\sqrt{2}-1|-\log |2-\sqrt{3}|=\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\)

Question 9.
\(\int_0^1 \frac{d x}{\sqrt{1-x^2}}\)
Solution:
Let I = \(\int_0^1 \frac{d x}{\sqrt{1-x^2}}=\left[\sin ^{-1} x\right]_0^1\)
= sin^-1 1 – sin^-1 0
= \(\frac{\pi}{2}\) – 0 = \(\frac{\pi}{2}\)

Question 10.
\(\int_0^1 \frac{d x}{1+x^2}\)
Solution:
Let I = \(\int_0^1 \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_0^1\)
= tan^-1 1 – tan^-1 0
= \(\frac{\pi}{4}\) – 0 = \(\frac{\pi}{4}\)

Question 11.
\(\int_2^3 \frac{d x}{x^2-1}\)
Solution:
Let I = \(\int_2^3 \frac{\mathrm{dx}}{\mathrm{x}^2-1}\)
= \(\left[\frac{1}{2} \log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|\right]_2^3\)
= \(\frac{1}{2}\left[\log \left|\frac{2}{4}\right|-\log \left|\frac{1}{3}\right|\right]\)
= \(\frac{1}{2}\left[\log \frac{1}{2}-\log \frac{1}{3}\right]\) = \(\frac{1}{2} \log \frac{3}{2}\)

Question 12.
\(\int_2^3 \frac{x d x}{x^2+1}\)
Solution:
Let I = \(\int_2^3 \frac{x}{x^2+1}\) dx
= \(\frac{1}{2} \int_2^3 \frac{2 x}{x^2+1}\) dx = \(\frac{1}{2}\left[\log \left(1+x^2\right)\right]_2^3\)
= \(\frac{1}{2}\)[log 10 – log 5] = \(\frac{1}{2}\) log \(\frac{10}{5}\)
= \(\frac{1}{2}\) log 2

Question 13.
\(\int_0^1 x e^{x^2} d x\)
Solution:
Let I = \(\int_0^1 x e^{x^2}\) dx
Put x² = t ⇒ 2x dx = dt ⇒ x dx = \(\frac{d t}{2}\)
L.L. x = 0 then t = 0 & U.L. x = 1 then t = 1
∴ I = \(\frac{1}{2} \int_0^1 e^t d t=\frac{1}{2}\left[e^t\right]_0^1\)
= \(\frac{1}{2}\) (e – 1)

Question 14.
\(\int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right)\) dx
Solution:
Let I = \(\int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right)\) dx
= \(2 \int_0^{\frac{\pi}{4}} \sec ^2 x d x+\int_0^{\frac{\pi}{4}} x^3 d x+2 \int_0^{\frac{\pi}{4}} 1 d x\)
= \(2(\tan x)_0^{\pi / 4}+\frac{1}{4}\left(x^4\right)_0^{\pi / 4}+2(x)_0^{\pi / 4}\)
= \(2 \tan \frac{\pi}{4}+\frac{1}{4}\left(\frac{\pi}{4}\right)^4+2\left(\frac{\pi}{4}\right)=2+\frac{\pi}{2}+\frac{\pi^4}{1024}\)

Question 15.
\(\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x\)
Solution:
Let I = \(\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right)\) dx
= –\(\int_0^\pi\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)\) dx
= \(-\int_0^\pi \cos x\) d x
= \(-[\sin x]_0^\pi\)
= -(sin π – sin 0) = 0

Question 16.
\(\int_0^{\frac{\pi}{2}} \cos ^2 x \)dx
Solution:
Let I = \(\int_0^{\pi / 2}\left(\frac{\cos 2 x+1}{2}\right)\) dx
= \(\frac{1}{2}\left[\left(\frac{\sin 2 x}{2}\right)_0^{\pi / 2}+(x)_0^{\pi / 2}\right]\)
= \(\frac{1}{2}\)[sin π – sin 0 + \(\frac{\pi}{2}\) – 0]
= \(\frac{1}{2}\)(0 – 0 + \(\)) = \(\frac{\pi}{4}\)

Question 17.
\(\int \frac{1}{\sqrt{1+x}-\sqrt{x}} d x\)
Solution:
Let I = \(\int \frac{1}{\sqrt{1+x}-\sqrt{x}} d x\)
= \(\int_0^1(\sqrt{1+x}+\sqrt{x}) d x=\left(\frac{(1+x)^{3 / 2}}{3 / 2}\right)_0^1+\left(\frac{x^{3 / 2}}{3 / 2}\right)_0^1\)
= \(\frac{2}{3}\) [2^3/2 – 1 + (1 – 0)]
= \(\frac{2}{3}\) (2^3/2) = \(\frac{2}{3}(2 \sqrt{2})\)
= \(\frac{4 \sqrt{2}}{3}\)

Question 18.
\(\int_0^{\frac{\pi}{2}} \sin ^3 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \sin ^3 x d x\)
I = \(\int_0^{\pi / 2}\left(\frac{3 \sin x-\sin 3 x}{4}\right) d x=\frac{1}{4}\left[\int_0^{\pi / 2} 3 \sin x d x-\int_0^{\pi / 2} \sin 3 x d x\right]\)
= \(\frac{1}{4}\left(-3 \cos x+\frac{\cos 3 x}{3}\right)_0^{\pi / 2}\)
= \(\frac{1}{4}\left(3-\frac{1}{3}\right)\) = \(\frac{2}{3}\)

II. Evaluate the following definite integrals

Question 1.
\(\int_0^1 \frac{2 x+3}{5 x^2+1}\) dx
Solution:

Question 2.
Evaluate \(\int_0^2 \frac{6 x+3}{x^2+4}\) dx
Solution:

Question 3.
Evaluate \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\) dx
Solution:
Let I = \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\) dx
I = \(\int_1^2\left[5-\frac{20 x+15}{x^2+4 x+3}\right]\) dx
Let \(\left(\frac{20 x+15}{x^2+4 x+3}\right)=\frac{A}{x+1}+\frac{B}{x+3}\)
20x + 15 = A(x + 3) + B(x + 1)
20 = A + B & 15 = 3A + B ⇒ A = -5/2 & B = 45/2
= \(5 \int_1^2 \mathrm{dx}+\frac{5}{2} \int_1^2 \frac{1}{\mathrm{x}+1} \mathrm{dx}-\frac{45}{2} \int_1^2 \frac{1}{\mathrm{x}+3} \mathrm{dx}\)
= \(\left(5 x+\frac{5}{2} \log (x+1)-\frac{45}{2} \log |x+3|\right)_1^2\)
= 10 + \(\frac{5}{2}\) log|3| – \(\frac{45}{2}\) log 5 – 5 – \(\frac{5}{2}\) log |2| + \(\frac{45}{2}\) log|4|
= 5 + \(\frac{5}{2}\) log|\(\frac{3}{2}\)| – \(\frac{45}{2}\) log|\(\frac{5}{4}\)|
= 5 – \(\frac{5}{2}\)(9 log \(\frac{5}{4}\) – log\(\frac{3}{2}\))

Question 4.
\(\int_0^1\left[x e^x+\sin \frac{\pi x}{4}\right]\) dx
Solution:
Let I = \(\int_0^1\left[x e^x+\sin \frac{\pi x}{4}\right]\) dx
= \(\int_0^1 x e^x d x+\int_0^1 \sin \frac{\pi x}{4} d x\) = \(\left(x e^x-e^x\right)_0^1-\frac{4}{\pi}\left(\cos \frac{\pi x}{4}\right)_0^1\)
= (e – e) – (0 – e⁰) – \(\frac{4}{\pi}\left(\cos \frac{\pi}{4}-\cos 0\right)\)
= 0 + 1 – \(\frac{4}{\pi}\left(\frac{1}{\sqrt{2}}-1\right)\)
= 1 – \(\frac{4}{\pi}\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right) \times \frac{\sqrt{2}}{\sqrt{2}}\) = 1 + \(\frac{2}{\pi}(2-\sqrt{2})\)
= 1 + \(\frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\)

Question 5.
\(\int_1^3 \frac{d x}{x^2(x+1)}\)
Solution:
Let, \(\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\) ……….. (1)
1 = Ax(x + 1) + B(x + 1) + C(x²) ………… (2)
Put x = 0, -1 on eqn (2) then
1 = B(0 + 1) ⇒ B = 1 & 1 = (-1)²C ⇒ C = 1
⇒ 1 = Ax² + Ax + Bx + B + Cx²
Comparing the coefficients of x² on both sides,
A + C = 0 ⇒ A = -C = -1
Substituting A, B, C values in (1) then
\(\frac{1}{x^2(x+1)}=\frac{-1}{x}+\frac{1}{x^2}+\frac{1}{(x+1)}\)
\(\int_1^3 \frac{1}{x^2(x+1)} d x=\int_1^3\left(\frac{-1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\right) d x\)
= \(\left(-\log |x|-\frac{1}{x}+\log |x+1|\right)_1^3=\left(\log \left|\frac{x+1}{x}\right|-\frac{1}{x}\right)_1^3\)
= \(\left(\log \left|\frac{4}{3}\right|-\frac{1}{3}\right)-\left(\log \left|\frac{2}{1}\right|-1\right)\)
= log \(\frac{4}{3}\) – log 2 + 1 – \(\frac{1}{3}\)
= log \(\frac{2}{3}\) + \(\frac{2}{3}\) = \(\frac{2}{3}\) + log\(\frac{2}{3}\)

Question 6.
\(\int_0^1 \frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}}\) dx
Solution:
Let I = \(\int_0^1 \frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}}\) dx
Put x = t² ⇒ dx = 2t dt ⇒ x^1/2 = t; x^1/4 = x^1/2