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Question Number 87839 by jagoll last updated on 06/Apr/20

$$\mathrm{I}=\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}\frac{\sin 4\mathrm{x}}{\cos {}^2\mathrm{x}\sqrt{\tan {}^4\mathrm{x}+1}}\mathrm{dx}$$

Answered by redmiiuser last updated on 06/Apr/20

$$\sqrt{\tan {}^{4}x+1}$$$$=\sqrt{\sin {}^{4}x+\cos {}^{4}x}/\cos {}^{2}x$$$$\frac{\sin 4x}{\sqrt{\sin {}^{4}x+\cos {}^{4}x}}$$$$=\frac{\sin 4x}{\sqrt{\frac{3+\cos 4x}{4}}}$$$${\int }_0^{\frac{\pi }{4}}\frac{2\sin 4x}{\sqrt{3+\cos 4x}}dx$$$$3+\cos 4x=z$$$$dz=4.\sin 4x.dx$$$${\int }_4^2\frac{2.dz}{4.\sqrt{z}}$$$$=\frac{1}{2}{\int }_4^2\frac{dz}{\sqrt{z}}$$$$={\left[\sqrt{z}\right]}_4^2$$$$=\sqrt{2}-2$$

Commented by jagoll last updated on 06/Apr/20

$$\mathrm{yes}\mathrm{sir}.\mathrm{thank}\mathrm{you}$$

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