Mathematics

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

$-2$ & $\dfrac{2}{3}$

SOLUTION
$\displaystyle\int \dfrac{dx}{\sqrt{\sin^3x\cdot \cos^5x}}=a\sqrt{\cot x}+b\sqrt{\tan^3x}+c$.......(1).
Now,
$\displaystyle\int \dfrac{dx}{\sqrt{\sin^3x\cdot \cos^5x}}$
$=\displaystyle\int \dfrac{cosec^2 xdx}{\sqrt{\cot x.\cos^4 x}}$ [ Dividing the numerator and the denominator by $\sin^2 x$]
$=\displaystyle\int \dfrac{cosec^2 x.\sec^2 xdx}{\sqrt{\cot x}}$
$=\displaystyle\int \dfrac{cosec^2 x.(1+\tan^2 x)dx}{\sqrt{\cot x}}$
$=\displaystyle\int \dfrac{cosec^2 x.dx}{\sqrt{\cot x}}$$+\displaystyle\int \dfrac{cosec^2 x.\tan^2 x\ dx}{\sqrt{\cot x}} =\displaystyle\int \dfrac{cosec^2 x.dx}{\sqrt{\cot x}}$$+\displaystyle\int \dfrac{\sec^2 x\ dx}{\sqrt{\cot x}}$
$=-\displaystyle\int \dfrac{d(\cot x)}{\sqrt{\cot x}}$$+\displaystyle\int {\sqrt{\tan x}d(\tan x)}$
$=-{2}\sqrt{\cot x}+\dfrac{2}{3}\sqrt{\tan^3x}+c$.
Comparing this with (1) we get,
$a=-2$ and $b=\dfrac{2}{3}$.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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