Integration by parts Examples

Find the integration of $ x \cos x d x $ using integration by parts.

Put $ f(x)=x $ (first function) and $ g(x)=\cos x( $ second function) Then, integration by parts gives
$ \int x \cos x d x=$$x \int \cos x d x-\int\left[\frac{d}{d x}(x) \int \cos x d x\right] d x $

$ =x \sin x-\int \sin x d x$$=x \sin x+\cos x+C $

Find the integration of $ e^{x} \sin x $ using integration by parts method.

Take $ e^{x} $ as the first function and $ \sin x $ as second function.
Then, Integrating by parts gives us,
$ \mathrm{I}=\int e^{x} \sin x d x $$=e^{x}(-\cos x)+\int e^{x} \cos x d x \\ =-e^{x} \cos x+\mathrm{I}_{1}(\text { say }) ....(1) $
Taking $ e^{x} $ and $ \cos x $ as the first and second functions, respectively, in $ \mathrm{I}_{1}, $ we get
$ \mathrm{I}_{1}=e^{x} \sin x-\int e^{x} \sin x d x $
Substituting the value of $ \mathrm{I}_{1} $ in $ (1), $ we get
$ \mathrm{I}=-e^{x} \cos x+e^{x} \sin x-\mathrm{I} $
or $ 2 \mathrm{I}=e^{x}(\sin x-\cos x) $
Finally, $ \mathrm{I}=\int e^{x} \sin x d x$$=\frac{e^{x}}{2}(\sin x-\cos x)+\mathrm{C} $

Find the integration of $ x^{2} \sin x d x $ using integration by parts method.

$ \int x^{2} \sin x d x$$=-x^{2} \cos x+\int 2 x \cos x d x $
$ =-x^{2} \cos x+2 x \sin x-\int 2 \sin x d x $
$ =-x^{2} \cos x+2 x \sin x+2 \cos x+C $

Find the integration of $ x^{3} e^{-x^{2}} d x $ using integration by parts method.

Firstly , we rewrite integral as $I= \int x^{2} (x e^{-x^{2}}) d x$
$ I=-\frac{1}{2} x^{2} e^{-x^{2}}-\int(-x) e^{-x^{2}} d x$
$=-\frac{1}{2} x^{2} e^{-x^{2}}-\frac{1}{2} e^{-x^{2}}+c $

Find the integration of $ e^{a x} \cos b x d x $ using integration by parts method.

Integrating by parts, taking $ e^{a x} $ as the first function, we find
$ I=e^{a x}\left(\frac{\sin b x}{b}\right)-\int a e^{a x}\left(\frac{\sin b x}{b}\right) d x $
Taking integration by parts second time$ I=e^{a x}\left(\frac{\sin b x}{b}\right)-a e^{a x}\left(\frac{-\cos b x}{b^{2}}\right)+\int a^{2} e^{a x}\left(\frac{-\cos b x}{b^{2}}\right) d x $integral on the RHS is just $ -a^{2} / b^{2} $ times the original integral $ I $ . Thus
$ I=e^{a x}\left(\frac{1}{b} \sin b x+\frac{a}{b^{2}} \cos b x\right)-\frac{a^{2}}{b^{2}} I $
Rearranging and adding integration constant gives
$ I=\frac{e^{a x}}{a^{2}+b^{2}}(b \sin b x+a \cos b x)+c $