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Question Number 186771 by mnjuly1970 last updated on 10/Feb/23

$$$$$$\mathrm{Q}:\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{integral}.$$$$$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{1}{1+{\sin }^4\left(x\right)+{\cos }^4\left(x\right)}\mathrm{d}x=?$$$$$$

Answered by Ar Brandon last updated on 10/Feb/23

$$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+{\sin }^{4}x+{\cos }^{4}x}={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{4}x}{{\sec }^{4}x+{\tan }^{4}x+1}dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{{\tan }^{2}x+1}{{({\tan }^{2}x+1)}^2+{\tan }^{4}x+1}d\left(\tan x\right)$$$$={\int }_0^{\mathrm{\infty }}\frac{t^2+1}{2t^4+2t^2+2}dt=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^2+1}{t^4+t^2+1}dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1+\frac{1}{t^2}}{t^2+1+\frac{1}{t^2}}dt=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1+\frac{1}{t^2}}{{(t-\frac{1}{t})}^2+3}dt$$$$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{du}{u^2+3}=\frac{1}{2\sqrt{3}}{\left[\arctan \left(\frac{u}{\sqrt{3}}\right)\right]}_{-\mathrm{\infty }}^{\mathrm{\infty }}$$$$=\frac{1}{2\sqrt{3}}(\frac{\pi }{2}--\frac{\pi }{2})=\frac{1}{2\sqrt{3}}(\frac{\pi }{2}+\frac{\pi }{2})=\frac{\pi }{2\sqrt{3}}$$

Commented by mnjuly1970 last updated on 11/Feb/23

$$thanksslotsir$$

Answered by integralmagic last updated on 10/Feb/23

$$=\frac{1}{2\sqrt{3}}arctan\left(\sqrt{3}/2tan2x\right){\mid }_0^{\pi /2}=\frac{\pi }{2\sqrt{3}}$$

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