Question

$\mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{integral}\int \frac{{(1-\cos x)}^{2/7}}{{(1+\cos x)}^{9/7}}dx\mathrm{is}$.

• A. $\frac{7}{11}{\left(\tan \frac{x}{2}\right)}^{\frac{11}{7}}+C$
• B. $\frac{7}{11}{\left(\sin \frac{x}{2}\right)}^{\frac{11}{7}}+C$
• C. $\frac{7}{11}{\left(\cos \frac{x}{2}\right)}^{\frac{11}{7}}+C$

Answer 1

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

To solve this integral we try simplifying the integrand into a particular trigonometric ratio.
Now, we know $\left(1-\cos x\right)=2{\sin }^2\frac{x}{2}\mathrm{and}(1+\cos x)=2{\cos }^2\frac{x}{2}$
Thus, substituting we get,
$I=\int \frac{{\left(1-\cos x\right)}^{2/7}}{{(1+\cos x)}^{9/7}}\mathrm{dx}\Rightarrow I=\int \frac{{\left(2{\sin }^2\frac{x}{2}\right)}^{2/7}}{{\left(2{\cos }^2\frac{x}{2}\right)}^{9/7}}\mathrm{dx}\Rightarrow I=\frac{1}{2}\int \frac{{\left(\sin \frac{x}{2}\right)}^{4/7}}{{\left(\cos \frac{x}{2}\right)}^{18/7}}\mathrm{dx}\Rightarrow I=\frac{1}{2}\int {\left(\tan \frac{x}{2}\right)}^{4/7}{\sec }^2\frac{x}{2}\mathrm{dx}\mathrm{Now},\mathrm{substituting}t=\tan \frac{x}{2},\mathrm{we}\mathrm{get}\mathrm{dt}=\frac{1}{2}{\sec }^2\frac{x}{2}\mathrm{dx}\mathrm{Thus},\mathrm{our}\mathrm{integral}\mathrm{becomes}:I=\int t^{\frac{4}{7}}\mathrm{dt}=\frac{7}{11}{\left(t\right)}^{\frac{11}{7}}+C\mathrm{Substituting}\mathrm{back}\mathrm{the}\mathrm{value}\mathrm{of}t,\mathrm{we}\mathrm{get}I=\frac{7}{11}{\left(\tan \frac{x}{2}\right)}^{\frac{11}{7}}+C$