Question Number 132414 by john_santu last updated on 14/Feb/21
Commented by john_santu last updated on 14/Feb/21
$$\underline{\mathrm{super}\:\mathrm{nice}\:\mathrm{integral}}\: \\ $$
Answered by liberty last updated on 14/Feb/21
![−∫_2 ^4 (((1−cos^2 x)^2 )/(cos x)) d(cos x)= −∫ ((1−2cos^2 x+cos^4 x)/(cos x)) d(cos x)= − {ln ∣cos x∣−cos^2 x+(1/4)cos^4 x }_2 ^4 = −{ln ∣((cos 4)/(cos 2))∣−(cos^2 (4)−cos^2 (2))+(1/4)[cos^4 (4)−cos^4 (2)]}](Q132436.png)
$$\:−\int_{\mathrm{2}} ^{\mathrm{4}} \:\frac{\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\mathrm{x}\right)^{\mathrm{2}} }{\mathrm{cos}\:\mathrm{x}}\:\mathrm{d}\left(\mathrm{cos}\:\mathrm{x}\right)= \\ $$$$−\int\:\frac{\mathrm{1}−\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\:\mathrm{d}\left(\mathrm{cos}\:\mathrm{x}\right)= \\ $$$$−\:\left\{\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{x}\mid−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}\:\right\}_{\mathrm{2}} ^{\mathrm{4}} = \\ $$$$−\left\{\mathrm{ln}\:\mid\frac{\mathrm{cos}\:\mathrm{4}}{\mathrm{cos}\:\mathrm{2}}\mid−\left(\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{4}\right)−\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2}\right)\right)+\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{cos}\:^{\mathrm{4}} \left(\mathrm{4}\right)−\mathrm{cos}\:^{\mathrm{4}} \left(\mathrm{2}\right)\right]\right\} \\ $$