Integration by parts challenge

We will solve this by using integration by parts twice.

Recall that ${\displaystyle \int udv=uv-\int vdu}$‍.

In this problem we will let $u=e^x$‍ and $dv=\sin (x)dx$‍. Then $du=e^{x}dx$‍ and ${\displaystyle v=\int \sin xdx=-\cos x}$‍.

Integration by parts gives:

To find ${\displaystyle \int e^x\cos xdx}$‍, we will integrate by parts again. This time, let $u=e^x$‍ and $dv=\cos xdx$‍. Then $du=e^{x}dx$‍ and $v=\sin x$‍.

Τώρα έχουμε:

Notice the like terms in the equation:

${\displaystyle \int e^x\sin xdx={\displaystyle -e^x\cos x+e^x\sin x-\int e^x\sin xdx}}$‍.

We can add ${\displaystyle \int e^x\sin xdx}$‍ to each side to obtain:

Dividing both sides by $2$‍ gives:

We will solve this by using integration by parts twice.

Recall that ${\displaystyle \int udv=uv-\int vdu}$‍.

In this problem we will let $u=(\ln x{)}^2$‍ and $dv=dx$‍. Then $du=2\ln x\cdot {\displaystyle \frac{1}{x}}dx$‍ and $v=x$‍.

Integration by parts gives:

To find ${\displaystyle \int \ln xdx}$‍, we will integrate by parts again. This time we will let $u=\ln x$‍ and $dv=dx$‍. Then $du={\displaystyle \frac{1}{x}}dx$‍ and $v=x$‍.

Τώρα έχουμε:

We will solve this by using integration by parts twice.

Recall that ${\displaystyle \int udv=uv-\int vdu}$‍.

In this problem we will let $u=x^2$‍ and $dv=\sin (\pi x)dx$‍. Then $du=2xdx$‍ and ${\displaystyle v=\int \sin (\pi x)dx=-{\displaystyle \frac{\cos (\pi x)}{\pi }}}$‍.

Integration by parts gives:

To find ${\displaystyle \int x\cos (\pi x)dx}$‍, we will integrate by parts again. This time we let $u=x$‍ and $dv=\cos (\pi x)dx$‍. Then $du=dx$‍ and $v={\displaystyle \int \cos (\pi x)dx={\displaystyle \frac{\sin (\pi x)}{\pi }}}$‍.

Τώρα έχουμε: