int frac{cos 2 x}{(sin x+cos x)^{2}} d x is | vedclass

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

Question

(B) Let $I = \int \frac{\cos 2x}{(\sin x + \cos x)^2} dx$. Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we have: $I = \int \frac{\cos^2 x - \sin

Vedclass · Accepted Answer

$\log |\sin x+\cos x|+C$

Vedclass · Answer

(B) Let $I = \int \frac{\cos 2x}{(\sin x + \cos x)^2} dx$. Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we have: $I = \int \frac{\cos^2 x - \sin^2 x}{(\sin x + \cos x)^2} dx$. Factorizing the numerator as a difference of squares: $I = \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\sin x + \cos x)^2} dx$. Simplifying the expression: $I = \int \frac{\cos x - \sin x}{\sin x + \cos x} dx$. Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$. Substituting these into the integral: $I = \int \frac{1}{t} dt = \log |t| + C$. Substituting back $t = \sin x + \cos x$: $I = \log |\sin x + \cos x| + C$. Thus,the correct option is $B$.

$\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$ is

Similar Questions

If $x \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right)$,then $\int \frac{\sin x - \cos x}{\sqrt{1 - \sin 2x}} e^{\sin x} \cos x \, dx = $

Evaluate the integral $\int \frac{\sqrt{\cot x}}{\sin x \cos x} d x = -f(x) + c$. Find $f(x)$.

Integrate the function: $\int \frac{3x}{1+2x^4} dx$

If $\int \frac{\log _e(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx = f(g(x)) + c$, then:

$\int \frac{y^2+\sqrt[3]{y^4}+\sqrt[6]{y^2}}{y\left(1+\sqrt[3]{y^2}\right)} d y=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module