Question

Given that \(\int e^{3x}\sin(x)\,dx=\frac{1}{10}e^{3x}(3\sin(x)-\cos(x))\). Evaluate \(\int e^{-3x}\cos(x)\,dx\)

Hint: \(\cos(x)=\sin(\frac{\pi}{2}-x)\) and \(\sin(x)=\cos(\frac{\pi}{2}-x)\)

Obviously substituting \(x\) by \(\frac{\pi}{2}-x\) won't work. I have no idea how to solve it.

Answer 1

The question specified that I must use \(\int e^{3x}\sin(x)\,dx=\frac{1}{10}e^{3x}(3\sin(x)-\cos(x))\) to solve \(\int e^{-3x}\cos(x)\,dx\). If not, I would have integrated it already.

Ping:@Callisto

Answer 2

well..substitute x by x- (pi/2) ..as far as i think,,this helps..

Answer 3

sorry,,pi/2 -x ..why you say it wont help ..

Answer 4

\[\large \int\limits e^{-3x}\cos x dx =- \int\limits e^{-3(\pi/2-u)}\cos(\pi/2-u)du\]\[\large =-\int\limits e^{-3\pi/2}e^{3u}\cos u du=-e^{-3\pi/2}\int\limits e^{3u}\cos u du\]apply the given formula

Answer 5

oops. it's sin u in the 2nd line, not cos u.

Answer 6

\[\large = -e^{3\pi/2}\left( \frac{ 1 }{ 10 }e^{3u}\right)\left( 3\sin u - \cos u \right)\]\[\large =-\frac{ 1 }{ 10 }e^{(-3\pi/2+3u)}(3\sin u - \cos u)\]\[\large =-\frac{ 1 }{ 10 }e^{-3(\pi/2-u)}(3\sin u - \cos u)\]\[\large = -e^{-3x}(3\sin (\pi/2 - x) - \cos (\pi/2 -x))\]\[\large =-e^{-3x}(\cos x - \sin x)\]

Answer 7

Ping @Callisto

Answer 8

@thomas5267 \[ x = \frac{\pi}{2}-u\]works

Answer 9

What do I do with this then?
\[
\begin{align*}
\int e^{3x}\sin(x)\,dx&=-\int e^{3\left(\frac{\pi}{2}-u\right)}\sin\left(\frac{\pi}{2}-u\right)\,du\\
&=-e^{\frac{3\pi}{2}}\int e^{-3u}\cos(u)\,du\\
\frac{1}{10e^\frac{3\pi}{2}}\left(\cos(x)-3\sin(x)\right)&=\int e^{-3u}\cos(u)\,du
\end{align*}
\]
The question is asking for \(\int e^{-3x}\cos(x)\,dx\), not \(\int e^{-3u}\cos(u)\,du\). Is my brain broken or what?

Answer 10

x is a dummy variable.

Answer 11

@thomas5267 you have some big problem, dude.

Answer 12

you replace u by pi/2 -x, and give it back so they are the same

Answer 13

Oh! Fxxx it. I get your point. @sirm3d
Myth confirmed: My brain is broken.

Answer 14

that was some mental block, i'd say. too much calculus, i think.

Answer 15

Thanks a lot! I think I am drinking too much coffee and doing too much calculus. You know, calculus and coffee don't mix.