int frac{1+cos 4 x}{cot x-tan x} d x is equal | vedclass

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

Question

(D) Let $I = \int \frac{1+\cos 4 x}{\cot x-	an x} d x$. Using the identity $1+\cos 4x = 2\cos^2 2x$ and $\cot x - 	an x = \frac{\cos x}{\sin x} - \f

Vedclass · Accepted Answer

$-\frac{1}{8} \cos 4 x+C$

Vedclass · Answer

(D) Let $I = \int \frac{1+\cos 4 x}{\cot x-	an x} d x$. Using the identity $1+\cos 4x = 2\cos^2 2x$ and $\cot x - 	an x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos 2x}{\frac{1}{2}\sin 2x} = 2\cot 2x$. Substituting these into the integral: $I = \int \frac{2\cos^2 2x}{2\cot 2x} d x = \int \frac{\cos^2 2x}{\frac{\cos 2x}{\sin 2x}} d x = \int \cos 2x \sin 2x d x$. Multiplying and dividing by $2$: $I = \frac{1}{2} \int 2\sin 2x \cos 2x d x = \frac{1}{2} \int \sin 4x d x$. Integrating $\sin 4x$: $I = \frac{1}{2} \left( -\frac{\cos 4x}{4} ight) + C = -\frac{1}{8} \cos 4x + C$.

$\int \frac{1+\cos 4 x}{\cot x-\tan x} d x$ is equal to

Similar Questions

$A$ minimum value of $\int_0^x t e^{t^2} d t$ is

If $\hat{i}+2 \hat{j}+3 \hat{k}$ and $3 \hat{i}+2 \hat{j}+\hat{k}$ are sides of a parallelogram, then a unit vector parallel to one of the diagonals of the parallelogram is

Two concentric hollow spherical shells have radii $r$ and $R$ $(R \gg r)$. $A$ charge $Q$ is distributed on them such that the surface charge densities are equal. The electric potential at the centre is

The electrochemical equivalent of a metal is $x \ g \ C^{-1}$. The equivalent weight of the metal is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module