Example 10.1.1 Evaluate $\ds\int \sin^5 x\,dx$. Rewrite the function: $$ \int \sin^5 x\,dx=\int \sin x \sin^4 x\,dx= \int \sin x (\sin^2 x)^2\,dx= \int \sin x (1-\cos^2 x)^2\,dx. $$ Now use $u=\cos x$, $du=-\sin x\,dx$: $$\eqalign{ \int \sin x (1-\cos^2 x)^2\,dx&=\int -(1-u^2)^2\,du\cr &=\int -(1-2u^2+u^4)\,du\cr &=-u+{2\over3}u^3-{1\over5}u^5+C\cr &=-\cos x+{2\over3}\cos^3 x-{1\over5}\cos^5x+C.\cr }$$ $\square$

Example 10.1.2 Evaluate $\ds\int \sin^6 x\,dx$. Use $\ds \sin^2x =(1-\cos(2x))/2$ to rewrite the function: $$\eqalign{ \int \sin^6 x\,dx=\int (\sin^2 x)^3\,dx&= \int {(1-\cos 2x)^3\over 8}\,dx\cr &={1\over 8}\int 1-3\cos 2x+3\cos^2 2x-\cos^3 2x\,dx.\cr} $$ Now we have four integrals to evaluate: $$\int 1\,dx=x$$ and $$\int -3\cos 2x\,dx = -{3\over 2}\sin 2x$$ are easy. The $\ds \cos^3 2x$ integral is like the previous example: $$\eqalign{ \int -\cos^3 2x\,dx&=\int -\cos 2x\cos^2 2x\,dx\cr &=\int -\cos 2x(1-\sin^2 2x)\,dx\cr &=\int -{1\over 2}(1-u^2)\,du\cr &=-{1\over 2}\left(u-{u^3\over 3}\right)\cr &=-{1\over 2}\left(\sin 2x-{\sin^3 2x\over 3}\right).} $$ And finally we use another trigonometric identity, $\ds \cos^2x=(1+\cos(2x))/2$: $$ \int 3\cos^2 2x\,dx=3\int {1+\cos 4x\over 2}\,dx= {3\over 2}\left(x+{\sin 4x\over 4}\right). $$ So at long last we get $$ \int \sin^6 x\,dx = {x\over8} -{3\over 16}\sin 2x -{1\over 16}\left(\sin 2x-{\sin^3 2x\over 3}\right) +{3\over 16}\left(x+{\sin 4x\over 4}\right)+C. $$ $\square$

Example 10.1.3 Evaluate $\ds\int\! \sin^2x\cos^2x\,dx$. Use the formulas $\ds \sin^2x =(1-\cos(2x))/2$ and $\ds \cos^2x =(1+\cos(2x))/2$ to get: $$ \int \sin^2x\cos^2x\,dx=\int {1-\cos(2x)\over2}\cdot {1+\cos(2x)\over2}\,dx. $$ The remainder is left as an exercise. $\square$