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 $I = \int \frac{1+\cos 4 x}{\cot x-	an x} dx$. Using the identity $1+\cos 4x = 2\cos^2 2x$ and $\cot x - 	an x = \frac{\cos x}{\sin x} - \fr

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int \frac{1+\cos 4 x}{\cot x-	an x} dx$. 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: $I = \int \frac{2\cos^2 2x}{2\cot 2x} dx = \int \frac{\cos^2 2x}{\frac{\cos 2x}{\sin 2x}} dx = \int \sin 2x \cos 2x dx$. Using the identity $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$: $I = \int \frac{1}{2} \sin 4x dx = \frac{1}{2} \left( -\frac{\cos 4x}{4} ight) + C = -\frac{1}{8} \cos 4x + C$.

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

Similar Questions

Find the following integral: $\int(ax^{2} + bx + c) dx$

If $\int \frac{1}{1-\cos x} dx = \tan \left(\frac{x}{\alpha} + \beta\right) + c$,then one of the values of $\frac{\pi \alpha}{4} - \beta$ is

$\int \frac{d x}{\sqrt{5+4 x-x^{2}}}=$

Integrate the function: $\sqrt{1-4x^2}$

$g(x)$ is an antiderivative of $f(x)=1+2^x \log 2$ and the graph of $y=g(x)$ passes through $\left(-1, \frac{1}{2}\right)$. Then the curve meets the $Y$-axis at

Vedclass Test Series

Exam Paper Generator

Online Exam Module