int frac{sin x-cos x}{sqrt{sin 2 x}} , dx is | vedclass

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

Question

(C) Let $I = \int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx$.We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2 \sin x \cos x

Vedclass · Accepted Answer

$-\log |\sin x+\cos x+\sqrt{\sin 2 x}|+C$

Vedclass · Answer

(C) Let $I = \int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx$.We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2 \sin x \cos x) = 1 - (\sin x - \cos x)^2$.Alternatively,we can write $\sin 2x = (\sin x + \cos x)^2 - 1$.So,$I = \int \frac{\sin x - \cos x}{\sqrt{(\sin x + \cos x)^2 - 1}} \, dx$.Let $t = \sin x + \cos x$.Then $dt = (\cos x - \sin x) \, dx$,which implies $(\sin x - \cos x) \, dx = -dt$.Substituting these into the integral:$I = \int \frac{-dt}{\sqrt{t^2 - 1}} = -\int \frac{dt}{\sqrt{t^2 - 1}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + C$,we get:$I = -\log |t + \sqrt{t^2 - 1}| + C$.Substituting $t = \sin x + \cos x$ back:$I = -\log |\sin x + \cos x + \sqrt{(\sin x + \cos x)^2 - 1}| + C$.Since $(\sin x + \cos x)^2 - 1 = 1 + \sin 2x - 1 = \sin 2x$,we have:$I = -\log |\sin x + \cos x + \sqrt{\sin 2x}| + C$.

$\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} \, dx$ is equal to

Similar Questions

Evaluate the integral: $\int \frac{x^4(x^{10}-1)}{x^{20}+3x^{10}+1} dx$

Integrate the function $\sqrt{\sin 2x} \cos 2x$.

$\int \sin \sqrt{x} \,d x=\ldots+C$ (where $C$ is a constant of integration.)

$\int \sqrt{4 \cos ^2 x - 5 \sin ^2 x} \cos x \, dx =$

$\int \cos \sqrt{x} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module