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Question Number 109457 by shahria14 last updated on 23/Aug/20

Answered by Dwaipayan Shikari last updated on 23/Aug/20

$$\int {sin}^{3}x(1-{cos}^{2}x)dx$$$$\int {sin}^{3}x-\int {sin}^3{xcos}^{2}x$$$$\int sinx(1-{cos}^{2}x)+\int {sin}^{2}x(-sinx){cos}^{2}xdx$$$$\int sinx-\int {sinxcos}^{2}x+\int (1-t^2)t^{2}dt$$$$-cosx+\frac{{cos}^{3}x}{3}+\frac{{cos}^{3}x}{3}-\frac{{cos}^{5}x}{5}+C$$

Answered by john santu last updated on 24/Aug/20

$$\frac{✓JS✓}{‘‘\bullet ‘‘}$$$$\int \sin {}^{4}x\sin xdx=\int {(1-\cos {}^{2}x)}^2(-d\left(\cos x\right))$$$$\int {(1-n^2)}^2(-dn)=-\int [n^4-2n^2+1]dn$$$$=-[\frac{1}{5}{n}^5-\frac{2}{3}{n}^3+n]+c$$$$=-\frac{\cos {}^{5}x}{5}+\frac{2\cos {}^{3}x}{3}-\cos x+c$$

Answered by 1549442205PVT last updated on 24/Aug/20

$$\int {\sin }^5\mathrm{xdx}=-\int {\sin }^4\mathrm{xd}\left(\mathrm{cosx}\right)=-\int {(1-{\cos }^2\mathrm{x})}^2\mathrm{d}\left(\mathrm{cosx}\right)$$$$=\int (-1+{2\cos }^2\mathrm{x}-{\cos }^4\mathrm{x})\mathrm{d}\left(\mathrm{cosx}\right)$$$$=-\mathbf{cosx}+\frac{2{\mathbf{cos}}^3\mathbf{x}}{3}-\frac{{\mathbf{cos}}^5\mathbf{x}}{5}+\mathbf{C}$$

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