Integration by Parts Questions

Dec 2008
509
2
Hi

I need help on the following question:

1)\(\displaystyle \int e^{x}sin(2x)dx\)
This is what i have done:
\(\displaystyle u = sin(2x) du = 2xcos(2x) dx\)

\(\displaystyle dv = e^{x}dx v = e^{x}\)

\(\displaystyle \int e^{x}sin(2x) = sin(2x)e^{x} - \int e^{x}2cos(2x)\)

\(\displaystyle =e^{x}sin(2x) - 2\int e^{x}cos(2x)\)

\(\displaystyle =e^{x}sin(2x) - 2\int cos(2x)e^{x} - \int -2e^{x}sin(2x)\)

This is where i am stuck, what should i do next??


2)\(\displaystyle \int arctan(x)dx\)
This is what i have done:
\(\displaystyle u=arctan(x) du = \frac{1}{1+x^2}dx\)

\(\displaystyle dv = dx v= x\)

\(\displaystyle \int arctan(x) dx = arctan(x)x - \int x * \frac{1}{(1+x^2))}\)

finally i get \(\displaystyle arctan(x)x - \frac{x^2}{2(1+x^2)}\)

P.S
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Hi

I need help on the following question:

1)\(\displaystyle \int e^{x}sin(2x)dx\)
This is what i have done:
\(\displaystyle u = sin(2x) du = 2xcos(2x) dx\)

\(\displaystyle dv = e^{x}dx v = e^{x}\)

\(\displaystyle \int e^{x}sin(2x) = sin(2x)e^{x} - \int e^{x}2cos(2x)\)

\(\displaystyle =e^{x}sin(2x) - 2\int e^{x}cos(2x)\)

\(\displaystyle =e^{x}sin(2x) - 2\int cos(2x)e^{x} - \int -2e^{x}sin(2x)\)

This is where i am stuck, what should i do next??


2)\(\displaystyle \int arctan(x)dx\)
This is what i have done:
\(\displaystyle u=arctan(x) du = \frac{1}{1+x^2}dx\)

\(\displaystyle dv = dx v= x\)

\(\displaystyle \int arctan(x) dx = arctan(x)x - \int x * \frac{1}{(1+x^2))}\)

finally i get \(\displaystyle arctan(x)x - \frac{x^2}{2(1+x^2)}\)

P.S
For number 1, you have the what you started with so added it back over to the LHS.
 
Oct 2008
1,035
295
2. Just in case a picture helps...

Your work so far in balloon sculpture (but without your final wrong turn)...



... where



... is the product rule - straight continuous lines differentiating downwards (integrating up) with respect to x.

The general drift for integration by parts being...



(This is a lazy way, doing without u and v.)



... where



... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (which is the inner function of the composite and hence subject to the chain rule).
_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus: Standard Integrals, Derivatives and Methods

Draw Balloon Calculus with LaTeX and Asymptote!
 
Last edited:
Dec 2008
509
2
For number 1, you have the what you started with so added it back over to the LHS.
After this i would still need the RHS?

\(\displaystyle -e^{x}sin(2x) = e^{x}sin(2x) - 2 \int cos(2x)e^{x}\)
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
After this i would still need the RHS?

\(\displaystyle -e^{x}sin(2x) = e^{x}sin(2x) - 2 \int cos(2x)e^{x}\)
From here
\(\displaystyle e^{x}sin(2x) - 2\int e^{x}cos(2x)\)

\(\displaystyle dv=e^x\rightarrow v=e^xdx\)
\(\displaystyle u=cos(2x)\rightarrow du=-2sin(2x)dx\)

\(\displaystyle e^{x}sin(2x)-2[vu-\int vdu]\)

Since you are adding the integral back over, you only have on the RHS

\(\displaystyle e^{x}sin(2x)-2vu\)
 
Last edited:
Dec 2008
509
2
i got: \(\displaystyle e^{x}sin(2x) - 2cos(2x)e^{x} + \int 4e^{x}sin(2x)\), is this the final answer?

If it is then this is incorrect according to the book's answer, which is
\(\displaystyle \frac{-e^{x}}{5}(2cos(2x)-sin(2x))+c\)
 
Jul 2007
894
298
New Orleans
i got: \(\displaystyle e^{x}sin(2x) - 2cos(2x)e^{x} + \int 4e^{x}sin(2x)\), is this the final answer?

If it is then this is incorrect according to the book's answer, which is
\(\displaystyle \frac{-e^{x}}{5}(2cos(2x)-sin(2x))+c\)
Here is another approach that I prefer since you do not have to deal with integration by parts.

Change everything to complex

\(\displaystyle \int e^{x}\sin{2x}dx\)

\(\displaystyle \int e^{x}e^{2ix}dx\)

\(\displaystyle \int e^{x+2ix}dx = \int e^{x(1+2i)}dx\)

\(\displaystyle \frac{e^{x(1+2i)}}{1+2i}\)

\(\displaystyle \bigg(\frac{e^{x(1+2i)}}{1+2i}\bigg)\bigg(\frac{1-2i}{1-2i}\bigg)\)

\(\displaystyle \frac{e^{x}e^{2ix}(1-2i)}{5}\)

\(\displaystyle \frac{e^{x}(\cos{2x}+i\sin{2x})(1-2i)}{5}\)

\(\displaystyle \frac{e^{x}(\cos{2x}-2i\cos{2x} +i\sin{2x}+2\sin{2x})}{5}\)

collect imaginary terms and you have your answer

\(\displaystyle \frac{e^{x}(-2\cos{2x} + \sin{2x})}{5} + C\)
 
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dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Number 1 in its entirety

\(\displaystyle \int e^xsin(2x)dx\)

\(\displaystyle u=sin(2x)\rightarrow du=2cos(2x)dx\)
\(\displaystyle dv=v=e^x\)

\(\displaystyle \int e^xsin(2x)dx=e^xsin(2x)-2\int e^xcos(2x)dx\)

\(\displaystyle u=cos(2x)\rightarrow du=-2sin(2x)dx\)
\(\displaystyle dv=v=e^x\)

\(\displaystyle \int e^xsin(2x)dx=e^xsin(2x)-2\bigg[e^xcos(2x)+2\int e^xsin(2x)dx\bigg]\)

\(\displaystyle 5\int e^xsin(2x)dx=e^xsin(2x)-2e^xcos(2x)+C\)

\(\displaystyle \int e^xsin(2x)dx=\frac{e^xsin(2x)-2e^xcos(2x)+C}{5}\)
 
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Apr 2010
384
153
Canada
Here is another approach that I prefer since you do not have to deal with integration by parts.

Change everything to complex

\(\displaystyle \int e^{x}\sin{2x}dx\)

\(\displaystyle \int e^{x}e^{2ix}dx\)

\(\displaystyle \int e^{x+2ix}dx = \int e^{x(1+2i)}dx\)

\(\displaystyle \frac{e^{x(1+2i)}}{1+2i}\)

\(\displaystyle \bigg(\frac{e^{x(1+2i)}}{1+2i}\bigg)\bigg(\frac{1-2i}{1-2i}\bigg)\)

\(\displaystyle \frac{e^{x}e^{2ix}(1-2i)}{5}\)

\(\displaystyle \frac{e^{x}(\cos{2x}+i\sin{2x})(1-2i)}{5}\)

\(\displaystyle \frac{e^{x}(\cos{2x}-2i\cos{2x} +i\sin{2x}+2\sin{2x})}{5}\)

collect imaginary terms and you have your answer

\(\displaystyle \frac{e^{x}(-2\cos{2x} + \sin{2x})}{5} + C\)
Interesting...but why do you feel that involving the complex plain is better then by parts?
 
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