Question

Evaluate $\int \left(sinx{)}^{-11/3}\right(cosx{)}^{-1/3}dx$

• A. $\frac{3}{8}[4co{t}^{2/3}x+co{t}^{8/3}x]+C$
• B. $\frac{-3}{8}[4ta{n}^{2/3}x+ta{n}^{8/3}x]+C$
• C. $\frac{3}{8}[4ta{n}^{2/3}x+ta{n}^{8/3}x]+C$
• D. $\frac{-3}{8}[4co{t}^{2/3}x+co{t}^{8/3}x]+C$

Answer 1

The correct option is D $\frac{-3}{8}[4co{t}^{2/3}x+co{t}^{8/3}x]+C$

$\int (sinx{)}^{-11/3}cos{x}^{-1/3}dx=\int \frac{1}{(si{n}^{11}xcosx{)}^{1/3}}dx$

$=\int \frac{1}{{(si{n}^{12}x\cdot \frac{cosx}{\sin x})}^{1/3}}dx$

$=\int \frac{cose{c}^{4}x}{(cotx{)}^{1/3}}dx=\int \frac{(1+co{t}^{2}x)cose{c}^{2}x}{(cotx{)}^{1/3}}dx$

[Substitute $cotx=t⟹-dt=cose{c}^{2}xdx$]

$=\int \frac{(1+t^2)}{(t{)}^{1/3}}dx$

$=-\int (t^{-1/3}+t^{5/3})dt$

$=-\left[\frac{t^{2/3}}{2/3}+\frac{t^{8/3}}{8/3}\right]+C=\frac{-3}{8}[4co{t}^{2/3}x+co{t}^{8/3}x]+C$