sin-4-x-cos-5-xdx-

Question Number 97322 by student work last updated on 07/Jun/20

$$\int\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}\centerdot\mathrm{cos}\:^{\mathrm{5}} \mathrm{xdx}=? \\ $$

Answered by bemath last updated on 08/Jun/20

∫ sin^4 x (1−sin^2 x)^2 cos x dx = ∫ q^4 (1−q^2 )^2 dq = ∫ q^4 (1−2q^2 +q^4 ) dq = ∫ q^4 −2q^6 +q^8 dq = (1/5)sin^5 x−(2/7)sin^7 x+(1/9)sin^9 x + c

$$\int\:\mathrm{sin}\:^{\mathrm{4}} {x}\:\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} \:\mathrm{cos}\:{x}\:{dx} \\ $$$$=\:\int\:{q}^{\mathrm{4}} \left(\mathrm{1}−{q}^{\mathrm{2}} \right)^{\mathrm{2}} \:{dq}\: \\ $$$$=\:\int\:{q}^{\mathrm{4}} \left(\mathrm{1}−\mathrm{2}{q}^{\mathrm{2}} +{q}^{\mathrm{4}} \right)\:{dq} \\ $$$$=\:\int\:{q}^{\mathrm{4}} −\mathrm{2}{q}^{\mathrm{6}} +{q}^{\mathrm{8}} \:{dq} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{5}}\mathrm{sin}\:^{\mathrm{5}} {x}−\frac{\mathrm{2}}{\mathrm{7}}\mathrm{sin}\:^{\mathrm{7}} {x}+\frac{\mathrm{1}}{\mathrm{9}}\mathrm{sin}\:^{\mathrm{9}} {x}\:+\:{c} \\ $$

Commented by Aziztisffola last updated on 07/Jun/20