Mathematics

# If $\displaystyle\int \frac{dx}{\sqrt{\sin^{3} x\cos^{5} x}}=a\sqrt{\cot x}+b\sqrt{\tan^{3} x}+C$, then

##### ANSWER

$a = -2,\>b = \displaystyle \frac {2}{3}$

##### SOLUTION
Let $\displaystyle I= \frac{dx}{\sqrt{\dfrac{\sin^{3} x}{\cos^{3} x}\cos^{8}x}}$

$\displaystyle=\int \frac{\sec^{4} x}{\sqrt{\tan^{3} x}}dx$

$\displaystyle=\int \frac{(1+\tan^{2} x)\sec^{2}x dx}{\sqrt{\tan^{3} x}}$

Put $\tan x =t$
$\Rightarrow \sec^{2} {x}dx=dt$

So, $\displaystyle I=\int \frac{1+t^{2}}{t^{3/2}}dt$

$\displaystyle =\int{t^{-3/2}}dt+\int{t^{1/2}}dt$

$\displaystyle= -2t^{-1/2} + \frac{2}{3} t^{3/2}$

$\displaystyle=-2\sqrt{\cot x}+\frac{2}{3}\sqrt{\tan^{3} x}+C$

Hence, $\displaystyle -2\sqrt{\cot x}+\frac{2}{3}\sqrt{\tan^{3} x}+C=a\sqrt{\cot x}+b\sqrt{\tan^{3} x}+C$

$\Rightarrow\displaystyle a=-2, b=\frac{2}{3}$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
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