Let the given integral be, I = 1 x- x d x Putting x = 1- ^2 (x 2 ) 1+ ^2 (x 2 ) and x = 2 x 2 1+ ^2 x 2 I= d x 1- ^2 (x 2 ) 1+ ^2 (x 2 ) - 2 x 2 1+ ^2 x 2 = ^2 (x 2 ) d x 1- ^2 (x 2 )-2 (x 2 ) Let (x 2 )= t 1 2 ^2 (x 2 ) dx = dt ^2 (x 2 ) dx =2 dt I= 2 d t 1-t^2-2 t = -2 d t t^2+2 t-1 = -2 d t t^2+2 t+1-2 =- 2 d t (t+1)^2-(2)^2 = 2 d t (2)^2-(t-1)^2 =2 2 2 | 2+t+1 2-t-1 |+C =1 2 | 2+ x 2 +1 2- x 2 -1 |+C

Evaluate: 1 x- x d x

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Evaluate: $\int \frac{1}{\cos x-\sin x} d x$

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Answer

Let the given integral be,
$ I =\int \frac{1}{\cos x-\sin x} d x$
$\text { Putting } \cos x =\frac{1-\tan ^2\left(\frac{x}{2}\right)}{1+\tan ^2\left(\frac{x}{2}\right)} \text { and } \sin x =\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}$
$\Rightarrow I=\int \frac{d x}{\frac{1-\tan ^2\left(\frac{x}{2}\right)}{1+\tan ^2\left(\frac{x}{2}\right)}-\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}$
$=\int \frac{\sec ^2\left(\frac{x}{2}\right) d x}{1-\tan ^2\left(\frac{x}{2}\right)-2 \tan \left(\frac{x}{2}\right)}$
$\text { Let } \tan \left(\frac{x}{2}\right)= t$
$\Rightarrow \frac{1}{2} \sec ^2\left(\frac{x}{2}\right) dx = dt$
$\sec ^2\left(\frac{x}{2}\right) dx =2 dt $
$\therefore I=\int \frac{2 d t}{1-t^2-2 t}$
$=\int \frac{-2 d t}{t^2+2 t-1}$
$=\int \frac{-2 d t}{t^2+2 t+1-2}$
$=-\int \frac{2 d t}{(t+1)^2-(\sqrt{2})^2}$
$=\int \frac{2 d t}{(\sqrt{2})^2-(t-1)^2}$
$=\frac{2}{2 \sqrt{2}} \ln \left|\frac{\sqrt{2}+t+1}{\sqrt{2}-t-1}\right|+C$
$=\frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2}+\tan \frac{x}{2}+1}{\sqrt{2}-\tan \frac{x}{2}-1}\right|+C$

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