Solveeit Logo
Login

Question

Question: \(\int_{}^{}\frac{\mathbf{\cos}\mathbf{x}\mathbf{-}\mathbf{\sin}\mathbf{x}}{\sqrt{\mathbf{\sin}\math...

∫cos⁡x−sin⁡xsin⁡2xdx\int_{}^{}\frac{\mathbf{\cos}\mathbf{x}\mathbf{-}\mathbf{\sin}\mathbf{x}}{\sqrt{\mathbf{\sin}\mathbf{2}\mathbf{x}}}\mathbf{dx}equals

A

cos⁡h−1(sin⁡x+cos⁡x)+c\cos h^{- 1}(\sin x + \cos x) + c

B

sin⁡h−1(sin⁡x+cos⁡x)+c\sin h^{- 1}(\sin x + \cos x) + c

C

−cos⁡h−1(sin⁡x+cos⁡x)+c- \cos h^{- 1}(\sin x + \cos x) + c

D

−sin⁡h−1(sin⁡x+cos⁡x)+c- \sin h^{- 1}(\sin x + \cos x) + c

Answer

cos⁡h−1(sin⁡x+cos⁡x)+c\cos h^{- 1}(\sin x + \cos x) + c

Explanation

Solution

I=∫cos⁡x−sin⁡xsin⁡2xdxI = \int \frac { \cos x - \sin x } { \sqrt { \sin 2 x } } d x

=∫cos⁡x−sin⁡x1+sin⁡2x−1dx= \int \frac { \cos x - \sin x } { \sqrt { 1 + \sin 2 x - 1 } } d x =∫cos⁡x−sin⁡x(cos⁡x+sin⁡x)2−1dx= \int_{}^{}{\frac{\cos x - \sin x}{\sqrt{(\cos x + \sin x)^{2} - 1}}dx}

Put cos⁡x+sin⁡x=t\cos x + \sin x = t ⇒(cos⁡x−sin⁡x)dx=dt\Rightarrow (\cos x - \sin x)dx = dt

⇒∫dtt2−1=cos⁡h−1t+c=cos⁡h−1(sin⁡x+cos⁡x)+c\Rightarrow \int \frac { d t } { \sqrt { t ^ { 2 } - 1 } } = \cos h ^ { - 1 } t + c = \cos h ^ { - 1 } ( \sin x + \cos x ) + c