Formula is:[tex] \int {sin^{n} x} \, dx =- \frac{1}{n}sin^{n-1}x cos x + \frac{n-1}{n} \int {sin^{n-2} x} \, dx [/tex]
We will use it here:[tex] \int {sin^{6} x} \, dx =- \frac{1}{6} sin^{5}xcosx+ \frac{5}{6} \int{sin^{4}x } \, dx [/tex]
And again:[tex] \int {sin^{4} x} \, dx =- \frac{1}{2} sin^{3} xcosx+ \frac{3}{4} \int{sin^{2}x } \, dx [/tex]
Finally:[tex] \int {sin^{2}x } \, dx=- \frac{1}{2}sinx cos x+ \frac{1}{2} [/tex]
The result:[tex]=- \frac{1}{6} sin^{5} xcosx- \frac{5}{24}sin^{3}xcosx- \frac{15}{48}sinxcosx+ \frac{15}{48}+C [/tex]
Thank you.
