Question

The value of the integral $\int \frac{(1-\cos \theta {)}^{2/7}}{(1+\cos \theta {)}^{9/7}}d\theta$ is

• A. $\frac{7}{11}{\left(\tan \frac{\theta }{2}\right)}^{\frac{11}{7}}+C$
• B. $\frac{7}{11}{\left(\cos \frac{\theta }{2}\right)}^{\frac{11}{7}}+C$
• C. $\frac{7}{11}{\left(\sin \frac{\theta }{2}\right)}^{\frac{11}{7}}+C$
• D. None of these

Answer 1

Answer: (A)

$\frac{7}{11}{\left(\tan \frac{\theta }{2}\right)}^{\frac{11}{7}}+C$

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

$1-\cos \theta =2{\sin }^2\theta$

& $1+\cos \theta =2{\cos }^2\theta$

$\int \frac{2^{2/7}{\sin }^{4/7}\theta /2}{2^{9/7}{\cos }^{18/7}\theta /2}d\theta$

$=\frac{1}{2}\int \frac{{\sin }^{4/7}\theta /2}{{\cos }^{4/7}\theta /2{\cos }^{184/7}\theta /2}d\theta$

$=\frac{1}{2}\int \frac{{\tan }^{4/7}\theta /2}{{\cos }^2\theta /2}d\theta$

$=\frac{1}{2}\int \left({\tan }^{4/7}\theta /2\right){\sec }^2\theta /2d\theta$

$=\frac{1}{2}\int {\tan }^{4/7}\theta /2d\left(\tan \theta /2\right)$

$=\frac{1}{2}\frac{{\tan }^2\theta /2}{11/7}=\frac{7{\tan }^{11/7}\theta /2}{11}+C$