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cos-2-x-tan-3-x-dx-




Question Number 121224 by liberty last updated on 06/Nov/20
    ∫ cos^2 x tan^3 x dx
$$\:\:\:\:\int\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}\:\mathrm{dx}\: \\ $$
Answered by benjo_mathlover last updated on 06/Nov/20
 ∫ ((sin^3 x)/(cos x)) dx = −∫ (((1−cos^2 x) d(cos x))/(cos x))   = −∫ (((1−u^2 ) du)/u)  = −(ln ∣u∣−(1/2)u^2 )+c   =−ln ∣cos x∣ +(1/2)cos^2 x + c
$$\:\int\:\frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx}\:=\:−\int\:\frac{\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)\:\mathrm{d}\left(\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{cos}\:\mathrm{x}} \\ $$$$\:=\:−\int\:\frac{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)\:\mathrm{du}}{\mathrm{u}} \\ $$$$=\:−\left(\mathrm{ln}\:\mid\mathrm{u}\mid−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{u}^{\mathrm{2}} \right)+\mathrm{c}\: \\ $$$$=−\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{x}\mid\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:+\:\mathrm{c} \\ $$

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