int frac{(1-cos x)^{2 / 7}}{(1+cos x)^{9 / 7} | vedclass

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

Question

(A) Let $I = \int \frac{(1-\cos x)^{2 / 7}}{(1+\cos x)^{9 / 7}} dx$. Using the identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \

Vedclass · Accepted Answer

$\frac{7}{11}\left(	an \frac{x}{2}ight)^{\frac{11}{7}}+C$

Vedclass · Answer

(A) Let $I = \int \frac{(1-\cos x)^{2 / 7}}{(1+\cos x)^{9 / 7}} dx$. Using the identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$, we get: $I = \int \frac{(2 \sin^2 \frac{x}{2})^{2/7}}{(2 \cos^2 \frac{x}{2})^{9/7}} dx$ $I = \int \frac{2^{2/7} \sin^{4/7} \frac{x}{2}}{2^{9/7} \cos^{18/7} \frac{x}{2}} dx$ $I = \int \frac{1}{2} \frac{\sin^{4/7} \frac{x}{2}}{\cos^{4/7} \frac{x}{2} \cdot \cos^2 \frac{x}{2}} dx$ $I = \int \frac{1}{2} 	an^{4/7} \frac{x}{2} \sec^2 \frac{x}{2} dx$. Let $t = 	an \frac{x}{2}$, then $dt = \frac{1}{2} \sec^2 \frac{x}{2} dx$. Substituting these into the integral: $I = \int t^{4/7} dt = \frac{t^{4/7 + 1}}{4/7 + 1} + C = \frac{t^{11/7}}{11/7} + C = \frac{7}{11} t^{11/7} + C$. Substituting back $t = 	an \frac{x}{2}$: $I = \frac{7}{11} \left(	an \frac{x}{2}ight)^{11/7} + C$.

$\int \frac{(1-\cos x)^{2 / 7}}{(1+\cos x)^{9 / 7}} d x=$

Similar Questions

$\int 3^{3^x} \cdot 3^x \, dx =$

$\int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx = $

The value of $\int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx$ is

$\int \frac{\sin x}{\sqrt{5 \sin ^2 x+6 \cos ^2 x}} \,d x=$

Evaluate the integral: $\int \frac{1}{\left(x+\frac{2}{x}\right) \sqrt{x^4+4 x^2+3}} d x$

Vedclass Test Series

Exam Paper Generator

Online Exam Module