calculus technique Definition and 1 Threads

$$\int_0^π \sin^ n x dx = \int_0^π \sin^ {n-1} x\sin x dx$$ Letting ##u=\sin^{n-1} x## and ##v^{'}= \sin x## then, $$\int_0^π \sin^ {n-1} x\sin x dx = [-\sin^ {n-1} x⋅\cos x]_0^π+ \int_0^π (n-1)\sin^{n-2} x ⋅ \cos^2 x dx$$ $$=\int_0^π...