Solution assignment 03 Integration by parts

Calculate:

$\displaystyle\int{\sin^2(x)}dx$

We can write this integral as:

$\displaystyle\int{\sin(x)}d(-\cos(x))$

Using integration by parts we get:

$\displaystyle\int{\sin(x)}d(-\cos(x))=$

$=-\sin(x)\cos(x)+\displaystyle\int{\cos(x)}d\sin(x))=$

$=-\sin(x)\cos(x)+\displaystyle\int{\cos^2(x)}dx$

We can rewrite the latter term by using a formula from trigonometry:

$\sin^2(x)+\cos^2(x)=1$

$\displaystyle\int{\cos^2(x)}dx=\displaystyle\int{(1-\sin^2(x))}dx=x-\displaystyle\int{\sin^2(x)}dx$

Now we have got:

$\displaystyle\int{\sin^2(x)}dx=-\sin(x)\cos(x)+x-\displaystyle\int{\sin^2(x)}dx$

and thus:

$\displaystyle\int{\sin^2(x)}dx=\displaystyle\frac{1}{2}[x-\sin(x)\cos(x)]+C$

Verify the result by differentiation.