Question

If $\int \frac{\cos x}{{\sin }^{3}x{(1+{\sin }^{6}x)}^{{\displaystyle \frac{2}{3}}}}dx=f\left(x\right){(1+{\sin }^{6}x)}^{\frac{1}{\lambda }}+c$ , where $c$ is constant of integration then $\lambda f\left(\frac{\pi }{3}\right)$

• A. $\frac{-9}{8}$
• B. $\frac{9}{8}$
• C. $2$
• D. $-2$

Answer 1

The correct option is D

$-2$

Explanation for correct option

Step1. Finding the value integration.

Given integration $\int \frac{\cos x}{{\sin }^{3}x{(1+{\sin }^{6}x)}^{{\displaystyle \frac{2}{3}}}}dx$

Consider,

$\begin{array}{rcl}\sin x& =& t\\ & \Rightarrow & \cos xdx=dt\end{array}$

Now, integration becomes

$\begin{array}{rcl}\int \frac{\cos xdx}{{\sin }^{3}x{(1+{\sin }^{6}x)}^{{\displaystyle \frac{2}{3}}}}& =& \int \frac{dt}{t^3{(1+t^6)}^{{\displaystyle \frac{2}{3}}}}\\ & =& \int \frac{dt}{t^7{(1+{\displaystyle \frac{1}{t^6}})}^{{\displaystyle \frac{2}{3}}}}\end{array}$

Again consider,
$\begin{array}{rcl}1+\frac{1}{t^6}& =& u\\ -6t^{-7}dt& =& du\end{array}$

Therefore,

$\begin{array}{rcl}\int \frac{dt}{t^7{(1+{\displaystyle \frac{1}{t^6}})}^{{\displaystyle \frac{2}{3}}}}& =& \frac{-1}{6}\int \frac{du}{u^{{\displaystyle \frac{2}{3}}}}\\ & =& \frac{-3}{6}{u}^{\frac{1}{3}}+c\\ & =& \frac{-1}{2}{u}^{\frac{1}{3}}+c\\ & =& \frac{-1}{2}{(1+\frac{1}{t^6})}^{\frac{1}{3}}+c\{from(ii\left)\right\}\\ & =& \frac{-1}{2}{(1+\frac{1}{{\sin }^{6}x})}^{\frac{1}{3}}+c\left\{from\right(i\left)\right\}\\ & =& \frac{-{(1+{\sin }^{6}x)}^{{\displaystyle \frac{1}{3}}}}{2{\sin }^{2}x}+c\end{array}$

Step2. Finding value of $\lambda f\left(\frac{\pi }{3}\right)$

Also given, $\int \frac{\cos x}{{\sin }^{3}x{(1+{\sin }^{6}x)}^{{\displaystyle \frac{2}{3}}}}dx=f\left(x\right){(1+{\sin }^{6}x)}^{\frac{1}{\lambda }}+c$

Therefore, $\begin{array}{rcl}\frac{-{(1+{\sin }^{6}x)}^{{\displaystyle \frac{1}{3}}}}{2{\sin }^{2}x}+c& =& f\left(x\right){(1+{\sin }^{6}x)}^{\frac{1}{\lambda }}+c\end{array}$

$\lambda =3\&f\left(x\right)=\frac{-1}{2{\sin }^{2}x}$

Now put the value in $\lambda f\left(\frac{\pi }{3}\right)$

$\begin{array}{rcl}\lambda f\left(\frac{\pi }{3}\right)& =& 3\frac{-1}{2\sin \left({\displaystyle \frac{\pi }{3}}\right)}\\ & =& \frac{-3}{2\times {\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}\\ & =& \frac{-3}{2\times {\displaystyle \frac{3}{4}}}\\ & =& -2\end{array}$

Hence, correct option is $\mathbf{\left(}\mathit{D}\mathbf{\right)}$