Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Q14 Find the integrals of the functions $\frac{\cos x - \sin x }{1+ \sin 2x }$

Answers (1)

best_answer

$\frac{\cos x-\sin x}{1+2\sin x}=\frac{\cos x-\sin x}{(sin^2x+cos^2x)+2 sin x.\cos x}$
$=\frac{\cos x-\sin x}{(\sin x+\cos x)^2}$

$\\sin x+\cos x =t\\ \therefore (\cos x-\sin x)dx = dt$

Now,
$=\int \frac{dt}{t^2}\\ =\int t^-2\ dt\\ =-t^{-1}+C\\ =-\frac{1}{(\sin x+\cos x)}+C$

Posted by

manish

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Single Girl Child Scholarship 2024: Registration extended till January 10; eligibility criteria, amount

anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

Latest Question