int frac{dx}{3 cos 2x + 5} equals | vedclass

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

Question

(C) Let $I = \int \frac{dx}{3 \cos 2x + 5}$.Using the identity $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$,we get:$I = \int \frac{dx}{3 \left(\frac{

Vedclass · Accepted Answer

$\frac{1}{4} 	an^{-1}\left(\frac{1}{2} 	an xight) + c$,where $c$ is the constant of integration.

Vedclass · Answer

(C) Let $I = \int \frac{dx}{3 \cos 2x + 5}$.Using the identity $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$,we get:$I = \int \frac{dx}{3 \left(\frac{1 - 	an^2 x}{1 + 	an^2 x}ight) + 5}$$I = \int \frac{1 + 	an^2 x}{3(1 - 	an^2 x) + 5(1 + 	an^2 x)} dx$$I = \int \frac{\sec^2 x}{3 - 3 	an^2 x + 5 + 5 	an^2 x} dx$$I = \int \frac{\sec^2 x}{8 + 2 	an^2 x} dx$Let $u = 	an x$,then $du = \sec^2 x dx$.$I = \int \frac{du}{8 + 2u^2} = \frac{1}{2} \int \frac{du}{4 + u^2}$Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}\left(\frac{x}{a}ight) + c$:$I = \frac{1}{2} \cdot \frac{1}{2} 	an^{-1}\left(\frac{u}{2}ight) + c$$I = \frac{1}{4} 	an^{-1}\left(\frac{	an x}{2}ight) + c$.

$\int \frac{dx}{3 \cos 2x + 5}$ equals

Similar Questions

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

For any integer $n \geq 2$,if $I_n = \int \cot^n x \, dx$,then $I_5 =$

If $I_1 = \int \frac{e^x}{e^{4x} + e^{2x} + 1} dx$ and $I_2 = \int \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} dx$,then $I_2 - I_1 =$

Let $\int \frac{2-\tan x}{3+\tan x} dx = \frac{1}{2}(\alpha x + \log_e |\beta \sin x + \gamma \cos x|) + C$,where $C$ is the constant of integration. Then $\alpha + \frac{\gamma}{\beta}$ is equal to:

If $\int \frac{5 \tan x}{\tan x-2} dx=ax+b \log |\sin x-2 \cos x|+c$,then $a+b=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module