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

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

Abhishek Panhale · Answer

1. substitute x=a.tan^2t, then solving u get a simplified expression as follows...........(after writing t again in terms of x)

= int [tan^-1sqrt(x/a)]dx.......................(1)

now put sqrt(x)=t

substitute value of dx in (1) n use byparts........

3. put x=sint..............this will help to get rid of roots.........

then u get an expression as,

1/2 int[sin^-1(x)]dx

use byparts..................

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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

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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

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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