# 1. sin-1 √[x/(x+a)] dx2. (sin-1 √x - cos-1 √x)/ ??(sin-1 √x + cos-1 √x) dx3. tan-1 {[√(1+x) - √(1-x)]/[√(1+x) + √(1-x)]} dx

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 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