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Question Number 124587 by mnjuly1970 last updated on 04/Dec/20

$$...nice◂::::▸calculus$$$$simplequestion::$$$$provethat::$$$${\int }_0^{\mathrm{\infty }}\frac{4}{\sqrt{4+x^4}}dx\stackrel{???}{=}{\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{sin\left(x\right)}}+{\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{cos\left(x\right)}}$$

Answered by mindispower last updated on 04/Dec/20

$$x=\sqrt{2}t$$$$=\int \frac{2\sqrt{2}dt}{\sqrt{1+t^4}}$$$$t=\sqrt{tg\left(x\right)}$$$$dt=\frac{1}{2{cos}^2\left(x\right)\sqrt{tg\left(x\right)}}$$$$=\frac{\sqrt{2}}{\sqrt{tg\left(x\right)}}.\frac{1}{cos\left(x\right)}dx={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{2}}{\sqrt{sin\left(x\right)cos\left(x\right)}}dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{2dx}{\sqrt{2sin\left(x\right)cos\left(x\right)}}2\int \frac{dx}{\sqrt{sin\left(2x\right)}}$$$$2x=w$$$$\Rightarrow ={\int }_0^{\pi }\frac{dw}{\sqrt{sin\left(w\right)}}={\int }_0^{\frac{\pi }{2}}\frac{dw}{\sqrt{sin\left(w\right)}}+{\int }_{\frac{\pi }{2}}^{\pi }\frac{dw}{\sqrt{sin\left(w\right)}}{\mid }_{w=\frac{\pi }{2}+x}$$$$={\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{sin\left(x\right)}}+{\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{cos\left(x\right)}}$$$$$$$$$$

Commented by mindispower last updated on 05/Dec/20

$$witheplesur$$

Commented by mnjuly1970 last updated on 04/Dec/20

$$excellent.sirmindspower$$$$asalways...$$

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