Question

Evaluate the following integral.

Answer 1

ANSWER :–

GIVEN FUNCTION :–

${ \bold{ = ∫ \frac{ \sin(x) + \cos( x ) }{ \sqrt{ \sin(2x) } } dx}}$

TO FIND :–

VALUE OF INTEGRATED FUNCTION...

${ \bold{ \green{ \huge{solution} : - }}}$

${ \bold{ = ∫ \frac{ \sin(x) + \cos(x) }{ \sqrt{ \sin(2x) } }dx }} \\ \\ \\ { \bold{ = ∫ \frac{ \sin(x) + \cos(x) }{ \sqrt{ - ( - \sin(2x) )} }dx }} \\ \\ \\ { \bold{ = ∫ \frac{ \sin(x) + \cos(x) }{ \sqrt{ - (1 - \sin(2x) - 1) } } dx}} \\ \\ \\ { \bold{ = ∫ \frac{ \sin(x) + \cos(x) }{ \sqrt{ - ( {sin}^{2} x + {cos}^{2} x - 2 \sin(x) \cos(x) - 1)}} dx}} \\ \\ \\ { \bold{ = ∫ \frac{ \sin(x) + \cos(x) }{ \sqrt{ - ( {( \sin(x) - \cos(x) )}^{2} - 1)} }dx }} \\ \\ \\ { \bold{ \: \: . \: \: now \: \: put \: \: \: \sin(x) - \cos(x) = t \: \: then - }} \\ \\ \\ { \bold{ \implies (\sin(x) + \cos(x)) dx = dt}} \\ \\ \\ { \bold{ = ∫ \frac{dt}{ \sqrt{ - ( {t}^{2} - 1)} } }} \\ \\ \\ { \bold{ = ∫ \frac{dt}{ \sqrt{1 - {t}^{2} } } }} \\ \\ \\ { \bold{ = { \sin }^{ - 1} (t) + c}} \\ \\ \\ { \bold{ \implies \boxed{ {ANSWER = {sin}^{ - 1} ( \sin(x) - cos(x)) + c}}}}$