∫cos⁡x−sin⁡xsin⁡2xdx equals:

I=∫cos⁡x−sin⁡xsin⁡2xdx=∫cos⁡x−sin⁡x1+sin⁡2x−1dx=∫cos⁡x−sin⁡x(cos⁡x+sin⁡x)2−1dxput cos⁡x+sin⁡x=t⇒ (cos⁡x−sin⁡x)dx=dt⇒ ∫dtt2−1=cos⁡h−1t+c =cos⁡h−1(sin⁡x+cos⁡x)+c.

∫cos⁡x−sin⁡xsin⁡2xdx equals:

I=∫cos⁡x−sin⁡xsin⁡2xdx=∫cos⁡x−sin⁡x1+sin⁡2x−1dx=∫cos⁡x−sin⁡x(cos⁡x+sin⁡x)2−1dxput cos⁡x+sin⁡x=t⇒ (cos⁡x−sin⁡x)dx=dt⇒ ∫dtt2−1=cos⁡h−1t+c =cos⁡h−1(sin⁡x+cos⁡x)+c.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\int \frac{\cos x - \sin x}{\sqrt{\sin 2 x}} d x$ equals:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

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

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

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$I = \int \frac{\cos x - \sin x}{\sqrt{\sin 2 x}} d x$

$\begin{array}{l} = \int \frac{\cos x - \sin x}{\sqrt{1 + \sin 2 x - 1}} d x \\ = \int \frac{\cos x - \sin x}{\sqrt{(\cos x + \sin x)^{2} - 1}} d x \end{array}$

put $\cos x + \sin x = t$

$\begin{array}{l} \Rightarrow (\cos x - \sin x) d x = d t \\ \Rightarrow \int \frac{d t}{\sqrt{t^{2} - 1}} = \cos h^{- 1} t + c \\ = \cos h^{- 1} (\sin x + \cos x) + c . \end{array}$