Question Number 87839 by jagoll last updated on 06/Apr/20
$$\mathrm{I}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\:\frac{\mathrm{sin}\:\mathrm{4x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\sqrt{\mathrm{tan}\:^{\mathrm{4}} \mathrm{x}+\mathrm{1}}}\:\mathrm{dx} \\ $$
Answered by redmiiuser last updated on 06/Apr/20
![(√(tan^4 x+1)) =(√(sin^4 x+cos^4 x))/cos^2 x ((sin 4x)/( (√(sin^4 x+cos^4 x)))) =((sin 4x)/( (√((3+cos 4x)/4)))) ∫_0 ^(π/4) ((2sin 4x)/( (√(3+cos 4x))))dx 3+cos 4x=z dz=4.sin 4x.dx ∫_4 ^2 ((2.dz)/(4.(√z))) =(1/2)∫_4 ^2 (dz/( (√z))) =[(√z)]_4 ^2 =(√2)−2](https://www.tinkutara.com/question/Q87851.png)
$$\sqrt{\mathrm{tan}\:^{\mathrm{4}} {x}+\mathrm{1}} \\ $$$$=\sqrt{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}/\mathrm{cos}\:^{\mathrm{2}} {x} \\ $$$$\frac{\mathrm{sin}\:\mathrm{4}{x}}{\:\sqrt{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{4}{x}}{\:\sqrt{\frac{\mathrm{3}+\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{4}}}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{2sin}\:\mathrm{4}{x}}{\:\sqrt{\mathrm{3}+\mathrm{cos}\:\mathrm{4}{x}}}{dx} \\ $$$$\mathrm{3}+\mathrm{cos}\:\mathrm{4}{x}={z} \\ $$$${dz}=\mathrm{4}.\mathrm{sin}\:\mathrm{4}{x}.{dx} \\ $$$$\int_{\mathrm{4}} ^{\mathrm{2}} \frac{\mathrm{2}.{dz}}{\mathrm{4}.\sqrt{{z}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{4}} ^{\mathrm{2}} \frac{{dz}}{\:\sqrt{{z}}} \\ $$$$=\left[\sqrt{{z}}\right]_{\mathrm{4}} ^{\mathrm{2}} \\ $$$$=\sqrt{\mathrm{2}}−\mathrm{2} \\ $$
Commented by jagoll last updated on 06/Apr/20
$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$