Question

integral (e^x(xcos x+sin x))dx=?

Answer 1

\[ \large \int\limits e^x(xcos x+\sin x)dx\]

Answer 2

I have been stuck on this question for sometime..I am missing something somewhere.

Answer 3

@ganeshie8 , @jim_thompson5910 Can you help?

Answer 4

its a case in integration of by parts!!

Answer 5

\[=\int\limits (x \cos x) e ^{x} dx+\int\limits (\sin x) e ^{x}dx\]
\[=x \cos x e ^{x}-\int\limits \left\{ x \left( -\sin x \right)+\cos x *1 \right\}e ^{x} dx+\[\sin x e ^{x}-\int\limits \cos x e ^{x}dx\]+c\]
there is some mistake,try to find out.

Answer 6

I am thinking of doing it in the form e^x(f(x)+f'(x)) By supplying terms and subtracting.

Answer 7

i have tried that also.

Answer 8

@surjithayer How did you get integral (xcos x)e^x dx=(xcos x)(e^x)

Answer 9

I am confused :/

Answer 10

you are doing that right and its integration is e^x.f(x)+c

Answer 11

@divu.mkr can you explain?

Answer 12

everything i wrote is gone

Answer 13

How??

Answer 14

it all ends up being in terms of:\[\int\limits_{}^{}e^x cosx dx= \frac{ 1 }{ 2 }e^x(sinx + cosx) + C\] and\[\int\limits_{}^{}e^x sinx dx= \frac{ 1 }{ 2 }e^x(sinx - cosx) + C\]. a rather simple integration by parts.

and seperating it like surji suggested and integrating only the first term by parts one gives:\[\frac{ x }{ 2 }e^x(sinx+cosx) - \frac{ 1 }{ 2 }\int\limits_{}^{} e^x (sinx + cosx)dx + C + \int\limits_{}^{}e^x sinx dx\]

Answer 15

was my own fault. i pressed a wrong key and then backspace

Answer 16

once**

Answer 17

Isnt it e^x (xcos x)?

Answer 18

the integral of that, by parts once, gives the first two terms of the last line.

taking u = x dv = e^xcosx

Answer 19

i wouldn't mind solving it entirely if you're not convinced :)

Answer 20

yeah please :P

Answer 21

Its just I want to to do this qn in form e^x((fx)+f'(x)) Format and not apply parts.

Answer 22

i'm not familiar with that method at all. only parts

Answer 23

did you want to see parts or you know how?

Answer 24

yeah I want to see parts.

Answer 25

i guess the only part is the:\[\int\limits_{}^{} e^x cosx dx\]
u = e^x; du = e^x dx
dv = cosx dx; v = sinx

\[\int\limits_{}^{} e^x cosx dx = e^xsinx - \int\limits_{}^{} e^x sinx dx\]
u = e^x; du = e^x dx
dv = sinx dx; v = -cosx

\[\int\limits_{}^{}e^x cosx dx= e^x sinx - \left( -e^x cosx + \int\limits_{}^{}e^x cosxdx \right)\] \[\int\limits_{}^{}e^x cosx dx = e^x(cosx + sinx) - \int\limits_{}^{}e^x cosx dx\]\[2\int\limits_{}^{}e^x cosx dx = e^x(sinx + cosx)\]\[\int\limits_{}^{}e^x cosx dx = \frac{ 1 }{ 2 }e^x (sinx + cosx)\]

Answer 26

do you have the answer..? i got one :D

Answer 27

yes final answer is 1/2e^x(xsin x+xcos x-cos x)+c

Answer 28

@Euler271 Yeah I get it now thanks!

Answer 29

glad i could help :)

Answer 30

i have tried that also.

Answer 31

Lots of integration by parts. Work on this in pieces (I've reused u and v several times but hopefully you can tell which is which by context):
$$
\int e^x(x\cos x+\sin x)dx\\
\int xe^x \cos x ~dx+\int xe^x \sin x~dx\\
\text{For }\int xe^x \cos x ~dx\\
\text{Let }u=x,dv=e^x\cos x\\du=x,v=\int e^x\cos x dx\\
\text{For }\int e^x\cos x dx\\
\text{Let }u=e^x,dv=\cos x\\
du=e^x,v=\sin x\\
\text{Then }uv-\int vdu=\\
e^x\sin x-\int e^x\sin x dx\\
\int e^x\sin x dx=\\
\text{Let }u=e^x,dv=\sin x\\
\text{then }du=e^x,v=-\cos x\\
uv-\int vdu = \\
-e^x\cos x+\int e^x\cos x dx\\
\text{Combining these results:}\\
\int e^x\cos x=\cfrac{e^x}{2}(\sin x+\cos x)\\
\text {So we now have }\\
\int xe^x \cos x ~dx\\
=\cfrac{xe^x}{2}(\sin x + \cos x)-\int \cfrac{e^x}{2}(\sin x + \cos x)dx+\\
\qquad \cfrac{e^x}{2}(\sin x - \cos x)\\
\text{Take, }\int\frac{e^x}{2}(\sin x + \cos x)dx\\
\int\frac{e^x}{2}\sin x + \int \cfrac{e^x}{2}\cos xdx\\
=\cfrac{e^x}{4}(\sin x -cos x)+\cfrac{e^x}{4}(\sin x +cos x)\\
=\cfrac{e^x}{2}\sin x\\
\text{So, }\\
\int xe^x \cos x ~dx\\
=\cfrac{xe^x}{2}(\sin x + \cos x)-\cfrac{e^x}{2}\sin x+\cfrac{e^x}{2}(\sin x - \cos x)\\
=\cfrac{e^x}{2}(x\sin x + (x-1)\cos x)\\
$$
Whew!!

That's it!