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
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