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Mathematics 37 Online
OpenStudy (anonymous):

Integration by Parts Just did an integration by parts question. I worked it out twice, with e^3x as u or sin(4x) as U, and got roughly the same answer, but neither time was it equal to what Woflramalpha gave. I was hoping to see if I did this question right; \[\int\limits_{}^{}e^{3x}\sin(4x)\] = \[e^{3x}(-1\sin(4x) + 3\cos(4x))\]

OpenStudy (anonymous):

When I used u = e^3x I got; \[\int\limits_{}^{}(e^{3x}(-1\sin(4x)+3\cos(4x)) / 7\]

OpenStudy (anonymous):

I can include any supporting work if needed to spot my error, but figured it would save me a lot of typing to compare answers.

OpenStudy (anonymous):

sin(4x)=u try this one..

OpenStudy (anonymous):

Cinar, that is the answer I gave in the original post. U = sin(4x)

OpenStudy (anonymous):

oh ok, but it works..

OpenStudy (anonymous):

you need to do that one more time and you will find e^3xsin4x take it other side and sum them..

OpenStudy (nenadmatematika):

\[U=e ^{3x}\]\[dU=3e ^{3x}\]\[dV=\sin4x\]\[V=(-1/4)\cos4x\] your integral is:\[(-1/4)e ^{3x}\cos4x+(3/4)\int\limits_{}^{}e ^{3x}\cos4xdx\] \[U=e ^{3x}\]\[dU=3e ^{3x}\]\[dV=\cos4xdx\]\[V=(1/4)\sin4x\]

OpenStudy (anonymous):

\[=(1/3)*e^{3x}\sin(4x) - (4/3)\int\limits_{}^{} e^{3x}\cos(4x)\] \[=(1/3)*e^{3x}\sin(4x) - (4/3)[ e^{3x}\cos(4x) - (4/3)\int\limits_{}^{}e^{3x}\sin(4x)\] Add -(4/3)e^{3x}sin(4x) to both sides, multiple both sides by -(3/1) and voila, final answer above.

OpenStudy (anonymous):

yeah that`s right..

OpenStudy (anonymous):

Right - I'm basically asking for someone to work the problem and confirm my answer. I usually use wolfram to confirm answers, but it is giving some whacky 1/25 using a weird formula. I know the steps I am taking are right, but I've arrived at three different answers (one for each u sub and one from wolfram).

OpenStudy (nenadmatematika):

so now your integral looks like this: \[(-1/4)e ^{3x}\cos4x+3/4((1/4)e ^{3x}\sin4x-3/4\int\limits_{}^{}e ^{3x}\sin4x))\]

OpenStudy (dumbcow):

are you sure you are using integration by parts correctly \[\int\limits u dv = uv -\int\limits_{?}^{?}v du\] \[u = e^{3x} ....... dv = \sin(4x)\] \[du = 3e^{3x} ........v = -\frac{1}{4}\cos(4x)\]

OpenStudy (anonymous):

http://integral-table.com/ take a look 118

OpenStudy (anonymous):

Yes, Dumbcow - I might have made it confusing by posting my answers to each U sub. I worked the problem twice, once with U = e^3x and once U = sin4x, to try to check my answer.

OpenStudy (dumbcow):

oh ok

OpenStudy (nenadmatematika):

\[\int\limits_{}^{}e ^{3x}\sin4x\]=some letter for example S....so now you have:\[\S=(-1/4)e ^{3x}\cos4x+(3/16)e ^{3x}\sin4x-(9/16)\S\]

OpenStudy (anonymous):

\[\frac{1}{25}e^{3x}(3\sin4x-4\cos4x)\]

OpenStudy (nenadmatematika):

S on one side and everything else on the other side:\[(25/16)\S=(-1/4)e ^{3x}\cos4x+(3/16)e ^{3x}\sin4x\]\[\S=(-4/25)e ^{3x}\cos4x+(3/25)e ^{3x}\sin4x\]

OpenStudy (nenadmatematika):

cinar and I have the same solution :D

OpenStudy (anonymous):

Damn, nice Nenadamatematika. I think my problem was not multiplying my fractions by the (1/4) after doing the integration. So I had (1/4), (3/4), and (3/4) for my fractions, which should have been (1/4), (3/16), and (9/16).

OpenStudy (anonymous):

Also I wish I could use that formula sheet Cinar :) thanks a lot guys!

OpenStudy (nenadmatematika):

you're welcome Arche :D

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