int frac{cos 2x}{(cos x + sin x)^2} , dx = | vedclass

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

Question

(C) We know that $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$.Substituting this into the integral:$\int \frac{(\cos x - \sin x

Vedclass · Accepted Answer

$\log (\cos x + \sin x) + c$

Vedclass · Answer

(C) We know that $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$.Substituting this into the integral:$\int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} \, dx = \int \frac{\cos x - \sin x}{\cos x + \sin x} \, dx$.Let $t = \cos x + \sin x$. Then $dt = (-\sin x + \cos x) \, dx = (\cos x - \sin x) \, dx$.The integral becomes $\int \frac{1}{t} \, dt = \log |t| + c$.Substituting back $t = \cos x + \sin x$,we get $\log |\cos x + \sin x| + c$.

$\int \frac{\cos 2x}{(\cos x + \sin x)^2} \, dx = $

Similar Questions

Let the equation of the curve passing through the point $(0,1)$ be given by $y=\int x^3 e^{x^4} d x$. If the equation of the curve is written in the form $x=f(y)$,then $f(y)=$

$\int(2 x-3) \sqrt{3 x+2} \, dx =$

$\int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x=$

$\int \frac{dx}{1+\sqrt{x}} = $

If $\int \frac{(2x+1)^6}{(3x+2)^8} dx = P \left( \frac{2x+1}{3x+2} \right)^Q + R$,then $\frac{P}{Q} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module