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1. sin -1 √[x/(x+a)] dx 2. (sin -1 √x - cos -1 √x)/ ??(sin -1 √x + cos -1 √x) dx 3. tan -1 {[√(1+x) - √(1-x)]/[√(1+x) + √(1-x)]} dx

1. sin-1 √[x/(x+a)] dx


2. (sin-1 √x - cos-1 √x)/ ??(sin-1 √x + cos-1 √x) dx


3. tan-1 {[√(1+x) - √(1-x)]/[√(1+x) + √(1-x)]} dx

Grade:11

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
1.
I = \int sin^{-1}\sqrt{\frac{x}{x+a}}dx
Integration by Parts
I = xsin^{-1}\sqrt{\frac{x}{x+a}}-\int \frac{1}{2}\sqrt{\frac{ax}{(x+a)^{2}}}dx
t = \sqrt{x}
dt = \frac{1}{2\sqrt{x}}dx
I = xsin^{-1}\sqrt{\frac{x}{x+a}}-\sqrt{a}\int\frac{t^{2}}{a+t^{2}}dt
I = xsin^{-1}\sqrt{\frac{x}{x+a}}-\sqrt{a}t +atan^{-1}(\frac{t}{\sqrt{a}}) +constant
I = xsin^{-1}\sqrt{\frac{x}{x+a}}-\sqrt{a}\sqrt{x}+atan^{-1}(\frac{\sqrt{x}}{\sqrt{a}}) +constant
2.
I = \int (\frac{sin^{-1}\sqrt{x}-cos^{-1}\sqrt{x}}{sin^{-1}\sqrt{x}+cos^{-1}\sqrt{x}})dx
Simply apply the substitution method here, we have
I = \frac{\sqrt{x(1-x)}+(x-1)sin^{-1}\sqrt{x}-xcos^{-1}\sqrt{x}}{sin^{-1}\sqrt{x}+cos^{-1}\sqrt{x}} + constant
3.
I = \int tan^{-1}(\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}})dx
Integration by parts
I = xtan^{-1}(\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}})-\frac{1}{2}\int\frac{x}{\sqrt{1-x^{2}}}dx
t = 1-x^{2}
dt = -2x.dx
I = xtan^{-1}(\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}})-\frac{1}{4}\int\frac{1}{\sqrt{t}}dt
I = xtan^{-1}(\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}})-\frac{\sqrt{t}}{2}+constant
I = xtan^{-1}(\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}})-\frac{\sqrt{1-x^{2}}}{2}+constant
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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