Question

Solve $\int \frac{co{s}^{9}x}{sinx}dx$

Answer 1

$\int \frac{{\cos }^{9}x}{\sin x}dx=\int \frac{{\cos }^{8}x\cos x}{\sin x}dx=\int \frac{{(1-{\sin }^{2}x)}^4\cos x}{\sin x}dx(\because {\cos }^{2}x+{\sin }^{2}x=1)$

Let $t=\sin xdt=\cos xdx$

$=\int \frac{{(1-t^2)}^4}{t}dt=\int \frac{{(1+t^4-2t^2)}^2}{t}dt=\int \frac{1+t^8+4t^4+2t^4-4t^2-4t^6}{t}dt=\int \frac{1}{t}dt+\int t^{7}dt+4\int t^{3}dt+2\int t^{3}dt-4\int tdt-4\int t^{5}dt=\int \frac{1}{t}dt+\int t^{7}dt+6\int t^{3}dt-4\int tdt-4\int t^{5}dt=\ln \left|t\right|+\frac{t^8}{8}+\frac{6t^4}{4}-\frac{4t^2}{2}-\frac{4t^6}{6}+C=\ln \left|t\right|+\frac{t^8}{8}+\frac{3t^4}{2}-2t^2-\frac{2}{3}{t}^6+C=\ln \left|\sin x\right|+\frac{{\sin }^{8}x}{8}+\frac{3{\sin }^{4}x}{2}-2{\sin }^{2}x-\frac{2}{3}{\sin }^{6}x+C$