Question

If $\int \frac{dx}{\sqrt{{\sin }^{3}x{\cos }^{5}x}}=a\sqrt{\cot x}+b\sqrt{{\tan }^{3}x}+c$ where c is an arbitrary constant of integration then the values of ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ are respectively :

• A. $-2$ & $\frac{2}{3}$
• B. $2$ & $-\frac{2}{3}$
• C. $2$ & $\frac{2}{3}$
• D. None of these

Answer 1

The correct option is A $-2$ & $\frac{2}{3}$

$\int \frac{dx}{\sqrt{{\sin }^{3}x\cdot {\cos }^{5}x}}=a\sqrt{\cot x}+b\sqrt{{\tan }^{3}x}+c$.......(1).

Now,

$\int \frac{dx}{\sqrt{{\sin }^{3}x\cdot {\cos }^{5}x}}$

$=\int \frac{cose{c}^{2}xdx}{\sqrt{\cot x.{\cos }^{4}x}}$ [ Dividing the numerator and the denominator by ${\sin }^{2}x$]

$=\int \frac{cose{c}^{2}x.{\sec }^{2}xdx}{\sqrt{\cot x}}$

$=\int \frac{cose{c}^{2}x.(1+{\tan }^{2}x)dx}{\sqrt{\cot x}}$

$=\int \frac{cose{c}^{2}x.dx}{\sqrt{\cot x}}$$+\int \frac{cose{c}^{2}x.{\tan }^{2}xdx}{\sqrt{\cot x}}$

$=\int \frac{cose{c}^{2}x.dx}{\sqrt{\cot x}}$$+\int \frac{{\sec }^{2}xdx}{\sqrt{\cot x}}$

$=-\int \frac{d\left(\cot x\right)}{\sqrt{\cot x}}$$+\int \sqrt{\tan x}d\left(\tan x\right)$

$=-2\sqrt{\cot x}+\frac{2}{3}\sqrt{{\tan }^{3}x}+c$.

Comparing this with (1) we get,

$a=-2$ and $b=\frac{2}{3}$.