Mathematics

The integral $$\displaystyle\int \sqrt{1+2\cot x(cosec x+\cot x)}dx\left(0 < x < \dfrac{\pi}{2}\right)$$ is equal to?


ANSWER

$$2log\left(\sin \dfrac{x}{2}\right)+c$$


SOLUTION
We have,
$$\displaystyle I= \int \sqrt{1+2cot x(cosec x+ cot x)}dx$$
We know that 
$$cosec  ^{2}x- cot^{2}x= 1$$
then,
$$\displaystyle I= \int \sqrt{cosec^{2}x- cot^{2}x+2 cotx(cosec x+ tan x)}dx$$
$$\displaystyle I= \int \sqrt{(cosec^{2}x- cot^{2}x+2 cotx\,cosec x+2 cot^2 x)}dx$$
$$\displaystyle I= \int (cosec^{2}x+cot^{2}x+2 cotx\, cosec x)^{\frac{1}{2}}dx$$
$$\displaystyle I= \int \sqrt{(cosec x+ cot x)^{2}}dx$$
$$\because (a+b)^{2}= a^{2}+b^{2}+2ab$$
$$\displaystyle I= \int (cosec x+ cot x)dx$$
$$\displaystyle I= \int (\frac{1}{sin x}+\frac{cos x}{sin x})dx$$
$$\displaystyle =\int (\frac{1+cos x}{sin x})dx$$
$$\displaystyle =\int \frac{1+2cos^{2}\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}dx$$
$$\displaystyle \because cos x= 2 cos^{2}\frac{x}{2}-1$$
$$\displaystyle \because sin x = 2sin\frac{x}{2}cos\frac{x}{2}$$
$$\displaystyle =\int \frac{2cos^{2}\frac{x}{2}}{2 sin\frac{x}{2}cos\frac{x}{2}}dx$$
$$\displaystyle =\int \frac{cos \frac{x}{2}}{sin \frac{x}{2}}dx$$
$$\displaystyle = cot \frac{x}{2}dx$$
on integrating and we get
$$\displaystyle I= log \left | \frac{sin\frac{x}{2}}{\frac{1}{2}} \right |+C$$
$$\displaystyle \therefore \int cot\,x\,dx= log\, sin x$$
$$\displaystyle I= 2 log(sin\frac{x}{2})+C$$
Hence, this is the answer.
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