Mathematics

The value of  $$\displaystyle\int_{0}^{\dfrac{\pi}{2}}{\dfrac{dx}{1+{\tan}^{3}{x}}}$$ is


SOLUTION
$$I=\displaystyle\int_{0}^{\dfrac{\pi}{2}}{\dfrac{dx}{1+{\tan}^{3}{x}}}$$          ......$$(1)$$

Replace $$x\rightarrow \dfrac{\pi}{2} - x$$

$$I=\displaystyle\int_{0}^{\dfrac{\pi}{2}}{\dfrac{dx}{1+{\tan}^{3}{\left(\dfrac{\pi}{2}-x\right)}}}$$

$$=\displaystyle\int_{0}^{\dfrac{\pi}{2}}{\dfrac{dx}{1+{\cot}^{3}{x}}}$$

$$=\displaystyle\int_{0}^{\dfrac{\pi}{2}}{\dfrac{dx}{1+\dfrac{1}{{\tan}^{3}{x}}}}$$  

$$=\displaystyle\int_{0}^{\dfrac{\pi}{2}}{\dfrac{{\tan}^{3}{x}dx}{{\tan}^{3}{x}+1}}$$        ......$$(2)$$

Adding $$(1)$$ and $$(2)$$ we get

$$\Rightarrow 2I =\displaystyle\int_{0}^{\dfrac{\pi}{2}}{\dfrac{1+{\tan}^{3}{x}}{1+{\tan}^{3}{x}}dx}$$

$$\Rightarrow 2I=\displaystyle\int_{0}^{\dfrac{\pi}{2}}{dx}$$

$$\Rightarrow 2I=\left[x\right]_{0}^{\dfrac{\pi}{2}}$$

$$\Rightarrow 2I=\dfrac{\pi}{2}$$

$$\therefore I=\dfrac{\pi}{4}$$
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