Question

Evaluate: $\int {\sin }^{7}xdx$

Answer 1

$\int {\sin }^{7}xdx$

$=\int {\sin }^{6}x\sin xdx$

$=\int {(1-{\cos }^{2}x)}^3\sin xdx$

$=\int (1-{\cos }^{6}x-3{\cos }^{2}x+3{\cos }^{4}x)\sin xdx$

$=\int \sin xdx-\int {\cos }^{6}x\sin xdx-3\int {\cos }^{2}x\sin xdx+3\int {\cos }^{4}x\sin xdx$

Let $t=\cos x\Rightarrow dt=-\sin xdx$

$=-\cos x+\int t^{6}dt+3\int t^{2}dt-3\int t^{4}dt$

$=-\cos x+\frac{t^7}{7}+3\frac{t^3}{3}-3\frac{t^5}{5}+c$

$=-\cos x+\frac{{\cos }^{7}x}{7}+3\frac{{\cos }^{3}x}{3}-3\frac{{\cos }^{5}x}{5}+c$

$=-\cos x+\frac{{\cos }^{7}x}{7}+{\cos }^{3}x-\frac{3{\cos }^{5}x}{5}+c$

$=\frac{{\cos }^{7}x}{7}-\frac{3{\cos }^{5}x}{5}+{\cos }^{3}x-\cos x+c$