Question

$\int si{n}^{-\frac{11}{3}}\left(x\right).co{s}^{-\frac{1}{3}}\left(x\right)dx$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

We can see that the given integral is also $\int si{n}^{m}x.co{s}^{n}xdx$ . Where m & n are rational numbers. So if $\frac{m+n-2}{2}$ is negative then we can either substitute cotx or tanx = t whichever is found suitable.

Here, $\frac{m+n-2}{2}=\frac{\left(\frac{-11}{3}\right)+\left(\frac{-1}{3}\right)-2}{2}=-3$

So, we can either substitute cotx = t or tanx = t

$\int si{n}^{\frac{-11}{3}}\left(x\right).co{s}^{\frac{-1}{3}}\left(x\right)dx$

Let’s rewrite the expression to get either tanx or cotx in it

$\int \frac{co{s}^{\frac{-1}{3}}\left(x\right)}{si{n}^{\frac{-1}{3}}\left(x\right).si{n}^4\left(x\right)}dxOr\int \frac{co{t}^{\frac{-1}{3}}\left(x\right)}{si{n}^4\left(x\right)}dxOr\int co{t}^{\frac{-1}{3}}\left(x\right)(cose{c}^{2}x{)}^{2}dxOr\int co{t}^{\frac{-1}{3}}\left(x\right)(1+co{t}^2(x\left)\right)\left(cose{c}^2\right(x\left)\right)dx$

Let’s substitute cot (x) = t

$-cose{c}^{2}x.dx=dt-\int t^{\frac{-1}{3}}.(1+t^2)dt=-\int t^{\frac{-1}{3}}.dt-\int t^{\frac{5}{3}}.dt-\frac{t^{\frac{2}{3}}}{\frac{2}{3}}-\frac{t^{\frac{8}{3}}}{\frac{8}{3}}+c$

Let’s substitute t = cot(x) in the above expression -

$-\frac{3.co{t}^{\frac{2}{3}}\left(x\right)}{2}-\frac{3.co{t}^{\frac{8}{3}}\left(x\right)}{8}+c$