Question

Evaluate $I=\int \frac{dx}{\sqrt[3]{3{\sin }^{11}x\cos x}}$

• A. $-\frac{3}{8}{t}^{-8/3}-\frac{3}{2}{t}^{-2/3}+c,$ where $(t=\tan x)$
• B. $\frac{3}{8}{t}^{-1/3}+\frac{3}{2}{t}^{-2/3}+c,$ where $(t=\tan x)$
• C. $\frac{1}{8}{t}^{-1/3}+\frac{5}{2}{t}^{-2/3}+c,$ where $(t=\tan x)$
• D. none of these

Answer 1

The correct option is A $-\frac{3}{8}{t}^{-8/3}-\frac{3}{2}{t}^{-2/3}+c,$ where $(t=\tan x)$

Here both the exponents $-\frac{11}{3}$ and $-\frac{1}{3}$ are negative numbers and their sum $(-\frac{11}{3}-\frac{1}{3})$ is $-4$.
Since, $4$ is an even number. Therefore, we put $\tan x=t\Rightarrow \frac{x}{{\cos }^{2}x}=dt$

Thus,
$I=\int \frac{dx}{{\sin }^{11/3}x{\cos }^{1/3}x}$
$=\int \frac{dx}{{\tan }^{11/3}x{\cos }^{4}x}$
$=\int \frac{(1+{\tan }^{2}x){\sec }^{2}xdx}{{\tan }^{11/3}x}$
$=\int \frac{(1+t^2)dt}{t^{11/3}}$
$=-\frac{3}{8}{t}^{-\frac{8}{3}}-\frac{3}{2}{t}^{-\frac{2}{3}}+c$
(where $t=\tan x$)