Question

Evaluate the indefinite integral of (e^(t/2))(2cos(t)dt)

Answer 1

looks like you'll be integrating by parts

Answer 2

How do I do that? I know that the integral of e^(t/2) is 2e^(t/2) but how do I do the other one? Can you help me with steps?

Answer 3

do you know how to do u-substitution?

Answer 4

\[\int e^{t/2}(2\cos t)~dt=2\int e^{t/2}~\cos t~dt=2\left[uv-\int v~du\right]\]
Let
\[\begin{matrix}
u=e^{t/2}&&&dv=\cos t~dt\\
du=\frac{1}{2}e^{t/2}~dt&&&v=\sin t
\end{matrix}\]
Integral changes to
\[2\left[e^{t/2}\sin t-\frac{1}{2} \int e^{t/2}\sin t~dt\right]\\
2e^{t/2}\sin t+\color{red}{\int e^{t/2}(-\sin t)~dt}\]
Integrate the red part by parts again, letting
\[\begin{matrix}
u=e^{t/2}&&&dv=-\sin t~dt\\
du=\frac{1}{2}e^{t/2}~dt&&&v=\cos t
\end{matrix}\]

Answer 5

and then you will have a term like the original one at the RHS, move it to LHS, you have 2* original one = something. divide both sides by 2, you have the answer. hehe... It takes long time to do.

Answer 6

Thank you.