Question

tan⁻¹√1-x/√1+x ,Integrate the given function defined on proper domain w.r.t. x.

Answer 1

HELLO DEAR,

put x = cost so that dx = -sint dt.

therefore, $\sf{\int}$ tan^{-1} √(1 - x)/√(1 + x) dx = $\sf{\int tan^{-1}x \sqrt{(1 - cost) \over (1 + cost)}(-sint)\,dt}$

$\sf{= \int tan^{-1}x{\sqrt{{2sin^2(t/2)\over 2cos^2(t/2)}}(-sint)\,dt}}$

$\sf{= \int[tan^{-1}x(tan{t \over 2}](-sint)\,dt = -1/2\int{t(sint)}\,dt}$

$\sf{= -1/2[t(-cost) - \int{1.(-cost)\,dt}]}$
[integrating by parts]

$\sf{=1/2tsint - 1/2siny + c}\\ \\ \sf{1/2x(cos^{-1}x) - 1/2\sqrt{1 - x^2} + c.}$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

we have to find the value of $\int{tan^{-1}\sqrt{\frac{1-x}{1+x}}}\,dx$


$\int{tan^{-1}\sqrt{\frac{1-x}{1+x}}}\,dx$

= $\int{tan^{-1}\sqrt{\frac{(1-x)^2}{(1-x^2}}}\,dx$

=$\int{tan^{-1}\frac{1-x}{\sqrt{1-x^2}}}\,dx$

can we assume $x = cos\theta$
differentiate with respect to x,
$dx=-sin\theta.d\theta$

= $\int{tan^{-1}\frac{1-cos\theta}{\sqrt{1-cos^2\theta}}}\,(-sin\theta.d\theta)$

=$\int{-sin\theta.tan^{-1}\frac{2sin^2\theta/2}{sin\theta}}\,d\theta$

=$\int{-sin\theta.tan^{-1}\frac{2sin^2\theta/2}{2sin\theta/2.cos\theta/2}}\,d\theta$

=$\int{-sin\theta.tan^{-1}(tan\theta/2)}\,d\theta$

=$-\int{sin\theta.\frac{\theta}{2}}\,d\theta$

=$-\frac{1}{2}\int{\theta.sin\theta}\,d\theta$
use integration by part ,

$=-\frac{1}{2}[\theta.-cos\theta+sin\theta]$

now, put $\theta=cos^{-1}x$

$=-\frac{1}{2}[-xcos^{-1}x+sin(cos^{-1}x)]$