Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{1}{1+{\cot }^{3}x}dx$ is equal to

(a) 0
(b) 1
(c) π/2
(d) π/4

Answer 1

(d) π/4

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+{\cot }^{3}x}dx.....\left(1\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+{\cot }^3\left({\displaystyle \frac{\mathrm{\pi }}{2}-x}\right)}dx \\ \therefore I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+{\tan }^{3}x}dx.....\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right)\mathrm{we}\mathrm{get} \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{1}{1+\mathrm{c}\mathrm{o}{\mathrm{t}}^{3}x}+\frac{1}{1+{\tan }^{3}x}\right]d\mathrm{x} \end{gathered}$$

$$\begin{gathered} ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{1+{\tan }^{3}x+1+\mathrm{co}{\mathrm{t}}^{3}x}{\left(1+\mathrm{c}\mathrm{o}{\mathrm{t}}^{3}x\right)\left(1+{\tan }^{3}x\right)}\right]\mathit{}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{2+{\tan }^{3}x+\mathrm{co}{\mathrm{t}}^{3}x}{1+{\tan }^{3}x+\mathrm{co}{\mathrm{t}}^{3}x+\mathrm{c}\mathrm{o}{\mathrm{t}}^{3}x{\tan }^{3}x}\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{2+{\tan }^{3}x+\mathrm{co}{\mathrm{t}}^{3}x}{1+{\tan }^{3}x+\mathrm{co}{\mathrm{t}}^{3}x+1}\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{2+{\tan }^{3}x+\mathrm{co}{\mathrm{t}}^{3}x}{2+{\tan }^{3}x+\mathrm{co}{\mathrm{t}}^{3}x}\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[1\right]dx \\ ={\left[\mathrm{x}\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =\frac{\mathrm{\pi }}{2} \\ \mathrm{Hence}I=\frac{\mathrm{\pi }}{4} \end{gathered}$$