Question

The value of 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

The correct option is A $\frac{7}{11}{\left(\tan \frac{\theta }{2}\right)}^{\frac{11}{7}}+c$

Let $I=\int \frac{(1-\cos \theta {)}^{2/7}}{(1+\cos \theta {)}^{9/7}}d\theta$
$=\int \frac{\left(2{\sin }^2\right(\theta /2){)}^{2/7}}{\left(2{\cos }^2\right(\theta /2){)}^{9/7}}d\theta$
$=\frac{1}{2}\int \frac{\left(\sin \right(\theta /2){)}^{4/7}}{\left(\cos \right(\theta /2){)}^{18/7}}d\theta$
Put $\frac{\theta }{2}=t\Rightarrow \frac{d\theta }{2}=dt$
$\therefore I=\int \frac{(\sin t{)}^{4/7}}{(\cos t{)}^{18/7}}dt$

$\Rightarrow I=\int (\tan t{)}^{4/7}{\sec }^{2}tdt$
Taking $\tan t=u\Rightarrow {\sec }^{2}tdt=du$
$\therefore I=\int u^{4/7}du=\frac{u^{11/7}}{11/7}+c$
$=\frac{7}{11}(\tan t{)}^{11/7}+c$
$=\frac{7}{11}{\left(\tan \frac{\theta }{2}\right)}^{11/7}+c$