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Answer 1 · 2018-02-05T13:12:21

# xcos^-1x-sqrt(1-x^2)+C.#

Recall that, #sqrt{(1-cos2theta)/(1+cos2theta)}=tantheta#.

So, if we subst. #x=cos2theta#, then, we have,

#tan^-1{(1-x)/(1+x)}=tan^-1{tan2theta}=2theta=cos^-1x#.

#:. I=inttan^-1{(1-x)/(1+x)}dx=intcos^-1xdx#,

#=int{(cos^-1x)*1}dx#.

Hence, Integrating by Parts, we get,

# I=(cos^-1x) int1dx-int[{d/dxcos^-1x}{int1dx}]dx#,

#=(cos^-1x)(x)-int[{-1/sqrt(1-x^2)}{x}]dx#,

#=xcos^-1x-int(-x/sqrt(1-x^2))dx#,

#=xcos^-1x-1/2int((-2x)/sqrt(1-x^2))dx#,

#=xcos^-1x-1/2int{(1-x^2)^(-1/2)*d/dx(1-x^2)}dx......(ast)#,

#=xcos^-1x-1/2*(1-x^2)^(-1/2+1)/(-1/2+1)#.

#rArr I=xcos^-1x-sqrt(1-x^2)+C.#

Note that, after #(ast)#, we have used the following Rule :

#int[f(x)]^n*f'(x)dx=[f(x)]^(n+1)/(n+1)+c, n ne -1#.