If x neq (2n+1) frac{pi}{2}, n in Z and cos x | vedclass

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

Question

(D) Given the integral $I = \int \left( \frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} \right)^2 dx$. Using trigonometric identities $\sin 2x = 2 \sin

Vedclass · Accepted Answer

$	an x - x + c$

Vedclass · Answer

(D) Given the integral $I = \int \left( \frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} ight)^2 dx$. Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = 2 \cos^2 x - 1$: $I = \int \left( \frac{\sin x + 2 \sin x \cos x}{1 + \cos x + 2 \cos^2 x - 1} ight)^2 dx$ $I = \int \left( \frac{\sin x (1 + 2 \cos x)}{\cos x (1 + 2 \cos x)} ight)^2 dx$ Since $\cos x 
eq -1/2$, we can cancel $(1 + 2 \cos x)$: $I = \int \left( \frac{\sin x}{\cos x} ight)^2 dx = \int 	an^2 x dx$ Using the identity $	an^2 x = \sec^2 x - 1$: $I = \int (\sec^2 x - 1) dx = 	an x - x + c$.

If $x \neq (2n+1) \frac{\pi}{2}, n \in Z$ and $\cos x \neq \frac{-1}{2}$, then evaluate the integral:
$\int \left( \frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} \right)^2 dx$

Similar Questions

The value of $\int \frac{x^{2}+1}{x^{2}-1} d x$ is

$\int \sqrt{\frac{1+x}{1-x}} \, dx = $ (where $C$ is a constant of integration.)

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

$\int \frac{x+\sin x}{1+\cos x} d x=$

$\int \frac{\sin 4x}{\sin x} \, dx =$ (where $C$ is a constant of integration.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If $x \neq (2n+1) \frac{\pi}{2}, n \in Z$ and $\cos x \neq \frac{-1}{2}$, then evaluate the integral:$\int \left( \frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} \right)^2 dx$