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 :


ANSWER

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


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
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