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Question Number 124178 by Dwaipayan Shikari last updated on 01/Dec/20

$${\int }_0^{\frac{\pi }{2}}\frac{1}{\sqrt{{cos}^{4}x+{sin}^{4}x}}dx$$

Commented by Dwaipayan Shikari last updated on 01/Dec/20

$$Ihavefound\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{4\sqrt{\pi }}$$

Commented by MJS_new last updated on 01/Dec/20

$$\mathrm{me}\mathrm{too}$$

Commented by Dwaipayan Shikari last updated on 01/Dec/20

https://brilliant.org/problems/another-apple-integral try this sir

Commented by MJS_new last updated on 01/Dec/20

$$p,q>0$$$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{dx}{\sqrt{p{\cos }^{4}x+q{\sin }^{4}x}}=$$$$[t=\frac{q^{1/4}}{p^{1/4}}\tan x\to dx=\frac{p^{1/4}}{q^{1/4}}{\cos }^{2}xdt]$$$$=\frac{1}{{\left(pq\right)}^{1/4}}\underset{0}{\stackrel{+\mathrm{\infty }}{\int }}\frac{dt}{\sqrt{t^4+1}}$$$$\mathrm{and}\mathrm{from}\mathrm{here}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}$$

Commented by Dwaipayan Shikari last updated on 01/Dec/20

$$DoyouhaveBrilliantapp,sir?$$

Commented by MJS_new last updated on 01/Dec/20

$$\mathrm{no}$$

Answered by mnjuly1970 last updated on 01/Dec/20

$$I={\underset{0}{\int }}^{\frac{\pi }{2}}\frac{1+{tan}^2\left(x\right)}{\sqrt{1+{tan}^4\left(x\right)}}dx$$$$\stackrel{tan\left(x\right)=t}{=}{\int }_0^{\mathrm{\infty }}\frac{dt}{\sqrt{1+t^4}}$$$$\stackrel{t^4=y}{=}\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{y^{-\frac{3}{4}}}{{(1+y)}^{\frac{1}{2}}}dy$$$$=\frac{1}{4}\beta (\frac{1}{4},\frac{1}{4})=\frac{1}{4}\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}$$$$=\frac{1}{4}\ast \frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{\sqrt{\pi }}✓$$$$$$

Commented by Dwaipayan Shikari last updated on 01/Dec/20

$$Thankingforconfirmation:)$$

Commented by mnjuly1970 last updated on 01/Dec/20

$$thankyousomuchsir$$$$Dwaipayan...$$

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