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Question Number 33331 by prof Abdo imad last updated on 14/Apr/18

$$calcilate{\int }_0^1\frac{dx}{{(1+x^2)}^3}$$

Commented by math khazana by abdo last updated on 19/Apr/18

$$letputI={\int }_0^1\frac{dx}{{(1+x^2)}^3}changementx=tantgive$$$$I={\int }_0^{\frac{\pi }{4}}\frac{1}{{(1+{tan}^{2}t)}^3}(1+{tan}^{2}t)dt$$$$={\int }_0^{\frac{\pi }{4}}\frac{dt}{{(1+{tan}^{2}t)}^2}={\int }_0^{\frac{\pi }{4}}{cos}^{4}tdt$$$$={\int }_0^{\frac{\pi }{4}}\frac{1}{4}{(1+cos\left(2t\right))}^{2}dt$$$$=\frac{1}{4}{\int }_0^{\frac{\pi }{4}}(1+2cos\left(2t\right)+{cos}^2\left(2t\right))dt$$$$=\frac{\pi }{16}+\frac{1}{2}{\int }_0^{\frac{\pi }{4}}cos\left(2t\right)+\frac{1}{8}{\int }_0^{\frac{\pi }{4}}(1+cos\left(4t\right))dt$$$$=\frac{\pi }{16}+\frac{1}{4}{\left[sin\left(2t\right)\right]}_0^{\frac{\pi }{4}}+\frac{\pi }{32}+\frac{1}{32}{\left[sin\left(4t\right)\right]}_0^{\frac{\pi }{4}}$$$$=\frac{3\pi }{32}+\frac{1}{4}+0\Rightarrow I=\frac{1}{4}+\frac{3\pi }{32}.$$

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