AP Inter 2nd Year Maths Exercise 7i Solutions

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

AP Inter 2nd Year Maths Integrals Solutions Exercise 7i

I.

Question 1.
Evaluate \(\int_0^1 \frac{x}{x^2+1} d x\)
Solution:
Put x² + 1 = t ⇒ 2xdx = dt
When x = 0 we have t = 1 and when x = 1 we have t = 2
∴ \(\int_0^1 \frac{\mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}=\frac{1}{2} \int_1^2 \frac{\mathrm{dt}}{\mathrm{t}}=\frac{1}{2}\left[\left.\log |\mathrm{t}|\right|_1 ^2=\frac{1}{2}[\log 2-\log 1]=\frac{1}{2} \log 2\right.\)

Question 2.
Evaluate \(\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} d x\)
Solution:
Put cosx = t ⇒ -sin xdx = dt
When x = 0 we have t = 1 and when x = \(\frac{\pi}{2}\) we have t = 0
⇒ \(\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} d x=-\int_1^0 \frac{d t}{1+t^2}=-\left[\tan ^{-1} t\right]_1^0=-\left[\tan ^{-1} 0-\tan ^{-1} 1\right]=-\left[-\frac{\pi}{4}\right]=\frac{\pi}{4}\)

Question 3.
Evaluate \(\int_{-1}^1 \frac{d x}{x^2+2 x+5}\)
Solution:
\(\int_{-1}^1 \frac{d x}{x^2+2 x+5}=\int_{-1}^1 \frac{d x}{\left(x^2+2 x+1\right)+4}=\int_{-1}^1 \frac{d x}{(x+1)^2+2^2}\)
Put x + 1 = t ⇒ dx = dt
When x = -1 we have t = 0 and when x = 1 we have t = 2
\(\int_{-1}^1 \frac{\mathrm{dx}}{(\mathrm{x}+1)^2+2^2}=\int_0^2 \frac{\mathrm{dt}}{\mathrm{t}^2+2^2}=\left[\frac{1}{2} \tan ^{-1} \frac{\mathrm{t}}{2}\right]_0^2=\frac{1}{2} \tan ^{-1} 1-\frac{1}{2} \tan ^{-1} 0=\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{8}\)

Question 4.
Evaluate \(\int_0^{\frac{\pi}{4}} 2 \tan ^3 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{4}} 2 \tan ^3 x d x\)

Question 5.
Evaluate \(\int_0^1 x e^x d x\)
Solution:
Let I = \(\int_0^1 x e^x d x\)
Let u = x and v = e^x and integrated by parts, we have
I = \(\left[x e^x\right]_0^1-\int_0^1\left[\left(\frac{d}{d x}(x)\right) \int e^x d x\right] d x=\left[x e^x\right]_0^1-\int_0^1 e^x d x=\left[x e^x\right]_0^1-\left[e^x\right]_0^1=\hat{e}-\hat{e}+1=1\)

Question 6.
Evaluate \(\int_1^2 \log x d x\)
Solution:
Let u = log x and v = 1, Now integrating by parts, we have
\(\int_1^2 \log x(1) d x=[\log x \cdot(x)]_1^2-\int_1^2(x) \frac{1}{x} d x\)
= [2 log2 – 1 log(1)] – \(\int_1^2 d x=2 \log 2-[x]_1^2=2 \log 2-[2-1]=2 \log 2-1\)

Question 7.
Evaluate \(\int_0^4 \frac{x^2}{1+x} d x\)
Solution:

II.

Question 1.
Evaluate \(\int_0^2 x \sqrt{x+2} d x\) (Put x + 2 = t²)
Solution:
\(\int_0^2 x \sqrt{x+2} d x\) Put x + 2 = t² ⇒ dx = 2tdt
When x = 0 we have t = \(\sqrt{2}\) and when x = 2 we have t = 2

Question 2.
Evaluate \(\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^5 \phi d \phi\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^5 \phi d \phi=\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^4 \phi \cos \phi d \phi\)
Put sin Φ = t ⇒ cosΦ dΦ = dt
When Φ = 0 we have t = 0 and when Φ = \(\frac{\pi}{2}\) we have t = 1
∴ I = \(\int_0^1 \sqrt{t}\left(1-t^2\right)^2 d t=\int_0^1 t^{\frac{1}{2}}\left(1+t^4-2 t^2\right) d t=\int_0^1\left[t^{\frac{1}{2}}+t^{\frac{9}{2}}-2 t^{\frac{5}{2}}\right] d t\)
= \(\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2 t^{\frac{7}{2}}}{\frac{7}{2}}\right]_0^1=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}=\frac{154+42-132}{231}=\frac{64}{231}\)

Question 3.
Evaluate \(\int_0^2 \frac{d x}{x+4-x^2}\)
Solution:
\(\int_0^2 \frac{\mathrm{dx}}{\mathrm{x}+4-\mathrm{x}^2}=\int_0^2 \frac{\mathrm{dx}}{-\left(\mathrm{x}^2-\mathrm{x}-4\right)}=\int_0^2 \frac{\mathrm{dx}}{-\left[\mathrm{x}^2-2 \cdot \mathrm{x} \frac{1}{2}+\frac{1}{4}-\frac{1}{4}-4\right]}\)
= \(\int_0^2 \frac{d x}{-\left[\left(x-\frac{1}{2}\right)^2-\frac{17}{4}\right]}=\int_0^2 \frac{d x}{\left(\frac{\sqrt{17}}{2}\right)^2-\left(x-\frac{1}{2}\right)^2}\)
Let x – \(\frac{1}{2}\) = t dx = dt when x = 0, t = \(-\frac{1}{2}\) and when x = 2, t = \(\frac{3}{2}\)

Question 4.
Evaluate \(\int_0^1 \sin ^{-1} x d x\)
Solution:
Let I = \(\int_0^1 \sin ^{-1} x d x \Rightarrow I=\int_0^1 \sin ^{-1} x \cdot 1 \cdot d x\)
Let u = sin^-1 x, v = 1 and integrated by parts, we get
I = \(\left[\sin ^{-1} x . x\right]_0^1-\int_0^1 \frac{1}{\sqrt{1-x^2}} x d x=\left[x \sin ^{-1} x\right]_0^1+\frac{1}{2} \int_0^1 \frac{(-2 x)}{\sqrt{1-x^2}} d x\)
Put 1 – x² = t ⇒ -2x dx = dt
When x = 0, t = 1 and when x = 1, t = 0
I = \(\left[x \sin ^{-1} x\right]_0^1+\frac{1}{2} \int_1^0 \frac{d t}{\sqrt{t}}=\left[x \sin ^{-1} x\right]_0^1+\frac{1}{2}[2 \sqrt{t}]_1^0=\sin ^{-1}(1)+[-\sqrt{1}]=\frac{\pi}{2}-1\)

Question 5.
Evaluate \(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x\)
Solution:
Let I = \(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x\)
Put x = tan θ ⇒ dx = sec² θdθ
When x = 0 we have θ = 0 and when x = 1 we have θ = \(\frac{\pi}{4}\)
∴ I = \(\int_0^{\frac{\pi}{4}} \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) \sec ^2 \theta \mathrm{~d} \theta=\int_0^{\frac{\pi}{4}} \sin ^{-1}(\sin 2 \theta) \sec ^2 \theta \mathrm{~d} \theta\)
= \(\int_0^{\frac{\pi}{4}} 2 \theta \sec ^2 \theta \mathrm{~d} \theta=2 \int_0^{\frac{\pi}{4}} \theta \sec ^2 \theta \mathrm{~d} \theta\)
Taking u = θ and v = sec² θ and integrating by parts, we get

Question 6.
Evaluate \(\int_0^1 x \tan ^{-1} x d x\)
Solution:
Applying the “By Parts Rule:, we have \(\int_0^1 \tan ^{-1} x(x) d x=\left[\tan ^{-1} x\left(\frac{x^2}{2}\right)\right]_0^1-\int_0^1\left(\frac{x^2}{2}\right) \frac{1}{1+x^2} d x\)