Inter 2nd Year Maths Exercise 7i Solutions

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

Inter 2nd Year Maths Integrals Solutions Exercise 7i

Integrals Exercise 7i Solutions

I. Evaluate the following definite integrals.

Question 1.
\(\int_0^1 \frac{x}{x^2+1}\) dx
Solution:
Let I = \(\int_0^1 \frac{x}{x^2+1}\) dx = \(\frac{1}{2} \int_0^1 \frac{2 x}{x^2+1}\) dx
= \(\frac{1}{2}\left(\log \left|x^2+1\right|\right)_0^1\)
= \(\frac{1}{2}\)(log 2 – log 1) = \(\frac{1}{2}\)log 2

Question 2.
\(\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x}\) dx
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x}\) dx
Put cos x = t ⇒ -sin x dx = dt
L.L : x = 0 we have t = 1 & U.L : x = \(\frac{\pi}{2}\) then t = 0
\(\int_1^0 \frac{1}{1+t^2}\) (-(dt)) = \(\int_0^1 \frac{1}{1+t^2}\) dt
= \(\left(\tan ^{-1} \mathrm{t}\right)_0^1=\frac{\pi}{4}\)

Question 3.
\(\int_{-1}^1 \frac{d x}{x^2+2 x+5}\)
Solution:
Let I = \(\int_{-1}^1 \frac{d x}{x^2+2 x+5}\) = \(\int_{-1}^1 \frac{d x}{x^2+2 x+1+4}\)
= \(\int_{-1}^1 \frac{\mathrm{dx}}{(\mathrm{x}+1)^2+2^2}\) = \(\frac{1}{2}\left(\tan ^{-1}\left(\frac{x+1}{2}\right)\right)_{-1}^1\)
= \(\frac{1}{2}\left(\tan ^{-1} \frac{2}{2}-\tan ^{-1} \frac{0}{2}\right)\) = \(\frac{\pi}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8}\)

Question 4.
\(\int_0^{\frac{\pi}{4}} 2 \tan ^3 x\) dx
Solution:

Question 5.
\(\int_0^1 x e^x\) dx
Solution:
Let I = \(\int_0^1 x e^x\) dx
= \(\left(x e^x\right)_0^1-\left(e^x\right)_0^1\)
= 1 e – 0 – e¹ + 1 = 1

Question 6.
\(\int_1^2 \log x\) dx
Solution:
Let I = \(\int_1^2 \log x\) dx = \(\int_1^2 \log x .1\) dx
= ∫ logx . dx = ∫[\(\frac{d}{d x}\)(log x) ∫1 dx]dx
= x log x – ∫\(\frac{1}{x}\)x dx = x log x – x
∴ \(\int_1^2 \log x\) dx = \((x \log x)_1^2-(x)_1^2\)
= 2 log 2 – 0 – (2 – 1)
= 2 log 2 – 1

Question 7.
\(\int_0^4 \frac{x^2}{1+x}\) dx
Solution:
Let I = \(\int_0^4 \frac{x^2}{1+x}\) dx = \(\int_0^4 \frac{x^2-1+1}{x+1}\) dx
= \(\int_0^4\left(x-1+\frac{1}{x+1}\right)\) dx = \(\left(\frac{x^2}{2}\right)_0^4-(x)_0^4+[\log (x+1)]_0^4\) dx
= 8 – 0 – 4 + log 5 – 0
= 4 + log 5

II. Evaluate the following definite integrals.

Question 1.
\(\int_0^2 x \sqrt{x+2}\) dx (Put x + 2 = t²)
Solution:

Question 2.
Evaluate \(\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^5 \phi d \phi\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^5 \phi d \phi\)
= \(\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi}\left(1-\sin ^2 \phi\right)^2 \cos \phi d \phi\)
Put sin Φ = t ⇒ cosΦ dΦ = dt
L.L Φ = 0, then t = 0, U.L : Φ = \(\frac{\pi}{2}\) then t = 1
∴ I = \(\int_0^1 \sqrt{\mathrm{t}}\left(1-\mathrm{t}^2\right)^2 \mathrm{dt}\) = \(\)
= \(\int_0^1\left(t^{9 / 2}+t^{1 / 2}-2 t^{51 / 2}\right)\) dt = \(\left(\frac{t^{11 / 2}}{\frac{11}{2}}+\frac{t^{3 / 2}}{\frac{3}{2}}-2 \frac{t^{7 / 2}}{\frac{7}{2}}\right)_0^1\)
= \(\frac{2}{11}\) + \(\frac{2}{3}\) – \(\frac{4}{7}\) = \(\frac{42+154-132}{231}\)
= \(\frac{64}{231}\)

Question 3.
\(\int_0^2 \frac{d x}{x+4-x^2}\)
Solution:

Question 4.
\(\int_0^1 \sin ^{-1} x d x\)
Solution:
Let I = \(\int_0^1 \sin ^{-1} x\) dx = \(\left(x \sin ^{-1} x\right)_0^1-\int_0^1 \frac{x}{\sqrt{1-x^2}}\) dx
Put 1 – x² = t ⇒ -2x dx = dt
L.L : x = 0 ⇒ t = 1 & U.L : x = 1 ⇒ t = 0
= (1) sin^-1(1) – 0 + \(\frac{1}{2} \int_0^1 \frac{d t}{\sqrt{t}}\) = \(\frac{\pi}{2}+\frac{1}{2}\left(\frac{t^{1 / 2}}{1 / 2}\right)_0^1\)
= \(\frac{\pi}{2}\) – 1

Question 5.
\(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\) dx
Solution:
Let I = \(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\) dx
Put x = tan θ ⇒ dx = sec² θ dθ ⇒ θ = tan^-1x
L.L : x = 0 then θ = 0 and & U.L: x = 1 then θ = \(\frac{\pi}{4}\)

Question 6.
\(\int_0^1 x \tan ^{-1} x\) dx
Solution: