Question

Heya✌
______
Solve this Integration
________________
Thanks for help☺
______________

Answer 1

$lets \: \: \\ \\ l = \frac{dx}{cosx - sinx} \\ \\ = > \frac{dx}{ \frac{1 - {tan}^{2} \frac{x}{2} }{1 + {tan}^{2} \frac{x}{2} } - \frac{2tan \frac{x}{2} }{1 + {tan}^{2} \frac{x}{2} } } \\ \\ = > \frac{ {sec}^{2} \frac{x}{2} dx }{1 - {tan}^{2} \frac{x}{2} - 2tan \frac{x}{2} } \\ \\put \: \: tan\frac{x}{2} = t \\ \\ = > \frac{1}{2} {sec}^{2} \frac{x}{2} .dx = dt \\ \\ = > {sec}^{2} \frac{x}{2} .dx = 2dt \\ \\ = > l = 2 \: \: intigrte \frac{dt}{1 - {t}^{2} - 2t } \\ \\ = > l = 2 \: \: intigrate \frac{dt}{ - ( {t}^{2} + 2t - 1)} \\ \\ = > l = - 2 \: \: intigrate \frac{dt}{ {t}^{2} + 2t + 1 - 2 } \\ \\ = > l = - 2 \: \: intigrate \frac{dt}{ {(t + 1)}^{2} - { \sqrt{2} }^{2} } \\ \\ = > l = 2 \: \: intigrate \frac{dt}{ { \sqrt{2} }^{2} - {(t + 1)}^{2} } \\ \\ = > l = 2 \times \frac{1}{2 \sqrt{2} } log \frac{ | \sqrt{2} + t + 1 | }{ | \sqrt{2} - t - 1 | } + c \\ \\ = > \frac{1}{ \sqrt{2} } log \frac{ | \sqrt{2} + tan \frac{x}{2} + 1 | }{ \sqrt{2} - tan \frac{x}{2} - 1} + c$

Answer 2

The answer is explained in the attachment.

Hope it helps!