MHB Integration by Parts: $\int u\cos(u)\,\mathrm{d}u$

[brunette15]

I have the following integral \int e^(2x) cos(e^x).

Let u = e^x

Do integration by parts:
\int u²cos(u) du = u²sin(u) - \int (2usin(u) du

Do integration by parts again for \int (2usin(u) du:

\int (2usin(u) du = -2ucos(u) - \int -2cos(u) du

Putting it all together:

\int e^(2x) cos(e^x) = e^(2x)sin(e^x) - 2e^xcos(e^x) + 2sin(e^x)

 

[Jameson]

Hi brunette15,

I don't think you u-substitution quite works, because $du=2e^{2x}$ that you haven't accounted for. Let's try another way.

Let $u=\cos(2x)$ and $dv = e^{2x}$. Then $$\int udv=uv-\int vdu$$.

What do you get when you try this?

Also, you might consider using $\LaTeX$. It takes a little time to learn but it makes your equations look very nice and clean. We can help you out if you are interested. :)

 

[brunette15]

Hi brunette15,

I don't think you u-substitution quite works, because $du=2e^{2x}$ that you haven't accounted for. Let's try another way.

Let $u=\cos(2x)$ and $dv = e^{2x}$. Then $$\int udv=uv-\int vdu$$.

What do you get when you try this?

Also, you might consider using $\LaTeX$. It takes a little time to learn but it makes your equations look very nice and clean. We can help you out if you are interested. :)

I see now! Thankyou! I have tried using latex however I am not too sure how to use it :/

 

[Prove It]

I have the following integral \int e^(2x) cos(e^x).

Let u = e^x

Do integration by parts:
\int u²cos(u) du = u²sin(u) - \int (2usin(u) du

Do integration by parts again for \int (2usin(u) du:

\int (2usin(u) du = -2ucos(u) - \int -2cos(u) du

Putting it all together:

\int e^(2x) cos(e^x) = e^(2x)sin(e^x) - 2e^xcos(e^x) + 2sin(e^x)

If your integral is $\displaystyle \begin{align*} \int{ \mathrm{e}^{2x}\cos{ \left( \mathrm{e}^x \right) } \,\mathrm{d}x} \end{align*}$, write it as $\displaystyle \begin{align*} \int{ \mathrm{e}^x\cos{ \left( \mathrm{e}^x \right) } \,\mathrm{e}^x\,\mathrm{d}x} \end{align*}$ and then let $\displaystyle \begin{align*} u = \mathrm{e}^x \implies \mathrm{d}u = \mathrm{e}^x\,\mathrm{d}x \end{align*}$, giving $\displaystyle \begin{align*} \int{ u\cos{ (u)}\,\mathrm{d}u} \end{align*}$. Now you can use Integration by Parts.