int frac{cos x - sin x}{1 + sin 2x} , dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We have the integral $I = \int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx$.Since $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \

Vedclass · Accepted Answer

$-\frac{1}{\cos x + \sin x} + c$

Vedclass · Answer

(A) We have the integral $I = \int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx$.Since $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$,the integral becomes:$I = \int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} \, dx$.Let $t = \sin x + \cos x$.Then $dt = (\cos x - \sin x) \, dx$.Substituting these into the integral,we get:$I = \int \frac{1}{t^2} \, dt = \int t^{-2} \, dt = \frac{t^{-1}}{-1} + c = -\frac{1}{t} + c$.Substituting back $t = \sin x + \cos x$,we get:$I = -\frac{1}{\sin x + \cos x} + c$.

$\int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx = $

Similar Questions

The value of $I=\int \frac{x^2}{(a+b x)^2} \,d x$ is

If $\int {\frac{{({x^2} - 1)\,dx}}{{({x^4} + 3{x^2} + 1)\,{{\tan }^{ - 1}}\left( {\frac{{{x^2} + 1}}{x}} \right)}}} = \ln | f(x) | + C$,then $f(x)$ is:

$\int \frac{dx}{e^{-2x}(e^{2x} + 1)^2} = $

If $\int 2^{2^{x}} \cdot 2^{x} \, dx = A \cdot 2^{2^{x}} + C,$ then $A$ is equal to

$\int \frac{\sec^8 x}{\operatorname{cosec} x} \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module