I=∫dxcosx cos2x =∫dxcosxcos2x-sin2x=∫dxcosxcos2x1-tan2x =∫sec2x dx1-tan2x put tan x=t sec2x dx=dt I=∫dt1-t2 =Sin-1t+C=Sin-1(tan 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

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

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

$C o s^{- 1} (\tan x) + c$

$S i n^{- 1} (\tan x) + c$

$T a n^{- 1} (\cos x) + c$

$C o s^{- 1} (\cos 2 x) + c$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$I = \int \frac{dx}{cosx \sqrt{\cos 2 x}} = \int \frac{dx}{cosx \sqrt{\cos^{2} x - \sin^{2} x}} = \int \frac{dx}{cosx \sqrt{\cos^{2} x [1 - \tan^{2} x]}} = \int \frac{\sec^{2} x dx}{\sqrt{1 - \tan^{2} x}} put \tan x = t \sec^{2} x dx = dt I = \int \frac{dt}{\sqrt{1 - t^{2}}} = S i n^{- 1} t + C = S i n^{- 1} (\tan x) + C$