Question

$\int \frac{\mathrm{dx}}{\sqrt[3]{3{\sin }^{11}x\mathrm{cosx}}}\mathrm{is}\mathrm{equal}\mathrm{to}$.

• A. $\frac{-3}{8}{\tan }^{-8/3}x-\frac{3}{2}{\tan }^{-2/3}x+C$
• B. $\frac{-3}{8}{\tan }^{-8/3}x+\frac{3}{2}{\tan }^{-2/3}x+C$
• C. $\frac{3}{8}{\tan }^{-8/3}x+\frac{3}{2}{\tan }^{-2/3}x+C$

Answer 1

The correct option is A $\frac{-3}{8}{\tan }^{-8/3}x-\frac{3}{2}{\tan }^{-2/3}x+C$

Here, both the exponents on sin and cosine $\left(-\frac{11}{3},-\frac{1}{3}\right)$ are negative and their sum is -4, which is an even number. Thus, we put $t=\mathrm{tanx};\Rightarrow \frac{\mathrm{dx}}{{\cos }^{2}x}=\mathrm{dt}$
$\mathrm{Now},I=\int \frac{\mathrm{dx}}{{\sin }^{11/3}x{\cos }^{1/3}x}\Rightarrow I=\int \frac{\mathrm{dx}}{\frac{{\sin }^{11/3}x}{{\cos }^{11/3}x}{\cos }^{11/3}x{\cos }^{1/3}x}\Rightarrow I=\int \frac{\mathrm{dx}}{{\tan }^{11/3}x{\cos }^{4}x}\Rightarrow I=\int \frac{{\sec }^{4}x\mathrm{dx}}{{\tan }^{11/3}x}\Rightarrow I=\int \frac{{\sec }^{2}x\left(1+{\tan }^{2}x\right)\mathrm{dx}}{{\tan }^{11/3}x}\mathrm{Now},\mathrm{substituting}t=\mathrm{tanx},\mathrm{we}\mathrm{get}\Rightarrow I=\int \frac{\left(1+t^2\right)\mathrm{dt}}{t^{11/3}}\Rightarrow I=\int \left(t^{-11/3}+t^{2-11/3}\right)\mathrm{dt}\Rightarrow I=\int \left(t^{-11/3}+t^{-5/3}\right)\mathrm{dt}\Rightarrow I=\frac{-3}{8}{t}^{-8/3}-\frac{3}{2}{t}^{-2/3}+C\mathrm{Substituting}\mathrm{back}t,\mathrm{we}\mathrm{get}:\Rightarrow I=\frac{-3}{8}{\tan }^{-8/3}x-\frac{3}{2}{\tan }^{-2/3}x+C$