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

$1.5 \ g$ of $CdCl_2$ was found to contain $0.9 \ g$ of $Cd$. Calculate the atomic weight of $Cd$.

The length of different internodes in a culm of sugarcane is variable because of

If $f: R \rightarrow R$ is defined by $f(x) = x - [x]$,where $[x]$ is the greatest integer not exceeding $x$,then the set of points of discontinuity of $f$ is

If $x=a\left[\cos \theta+\log \left\{\tan \left(\frac{\theta}{2}\right)\right\}\right]$ and $y=a \sin \theta$,then $\frac{dy}{dx}$ is equal to

If $2.2 \, kW$ power is transmitted through a $10 \, \Omega$ line at $22000 \, V$,the power loss in the form of heat will be ............. $W$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module