# easy integration by parts- cant figure out what im doing wrong

• Nov 12th 2009, 11:07 AM
twostep08
easy integration by parts- cant figure out what im doing wrong
ok its
(integral) sin x cos x dx
i did u= sinx dv= cosx du=cos x v=sinx

i then plug in and get sin^2 x - (integral)(sin x cos x)

that second term is exactly the problem we started with, so how does that help me?--it seems like i will just be doing an endless integration by continuing when that is clearly not what the answer dictates...
• Nov 12th 2009, 11:17 AM
Quote:

Originally Posted by twostep08
ok its
(integral) sin x cos x dx
i did u= sinx dv= cosx du=cos x v=sinx

i then plug in and get sin^2 x - (integral)(sin x cos x)

that second term is exactly the problem we started with, so how does that help me?--it seems like i will just be doing an endless integration by continuing when that is clearly not what the answer dictates...

post deleted,

sorry i messed up. i'll put the right answer in a moment.
• Nov 12th 2009, 11:20 AM
Soroban
Hello, twostep08!

Quote:

$\displaystyle \int \sin x\cos x\,dx$

i did: .$\displaystyle \begin{array}{ccccccc} u&=& \sin x& dv&=& \cos x \\ du &=& \cos x & v &=& \sin x\end{array}$

i then plug in and get: .$\displaystyle \sin^2\!x - \int \sin x\cos x\,dx$

That second term is exactly the problem we started with,
so how does that help me?

We have: .$\displaystyle I \;=\; \int\sin x\cos x\,dx$

$\displaystyle \text{Using By-parts, we get: }\;I =\;\sin^2\!x - \underbrace{\int\sin x\cos x\,dx}_{\text{This is }I} +\: c$

Hence: .$\displaystyle I \;=\;\sin^2\!x - I + c \quad\Rightarrow\quad 2I \;=\;\sin^2\!x + c \quad\Rightarrow\quad I \;=\;\tfrac{1}{2}\sin^2\!x + C$

Therefore: .$\displaystyle \int\sin x\cos x\,dx \;=\;\tfrac{1}{2}\sin^2\!x + C$

• Nov 12th 2009, 11:25 AM
e^(i*pi)
Quote:

Originally Posted by twostep08
ok its
(integral) sin x cos x dx
i did u= sinx dv= cosx du=cos x v=sinx

i then plug in and get sin^2 x - (integral)(sin x cos x)

that second term is exactly the problem we started with, so how does that help me?--it seems like i will just be doing an endless integration by continuing when that is clearly not what the answer dictates...

I'd use a trig sub for this one - no need for integration by parts

$\displaystyle sin(2x) = 2sin(x)cos(x) \: \rightarrow \: sin(x)cos(x) = \frac{1}{2}sin(2x)$

$\displaystyle \frac{1}{2} \int sin(2x)\,dx = -\frac{1}{4}cos(2x)+C$

Edit:

My answer is the same as Soroban's above. First let the constant of integration be $\displaystyle k$ to make it different from Soroban

$\displaystyle -\frac{1}{4}cos(2x)+k$

$\displaystyle cos(2x) = 1-2sin^2(x)$

$\displaystyle -\frac{1}{4}(1-2sin^2(x))+k$

$\displaystyle -\frac{1}{4} + \frac{1}{2}sin^2(x) + k$

As $\displaystyle -\frac{1}{4}$ is constant we can define the following $\displaystyle C = k - \frac{1}{4}$

Therefore $\displaystyle -\frac{1}{4}cos(2x)+k = \frac{1}{2}sin^2(x) + C$
• Nov 12th 2009, 11:41 AM
tom@ballooncalculus
It's worth knowing, though (edit: and this is just what Soroban is also pointing out), that when the 'endlessness' thing occurs in integration by parts, you can often break out of it like this...

http://www.ballooncalculus.org/asy/parts/sinCos.png

... i.e. by solving the top row for I.

http://www.ballooncalculus.org/asy/parts/sinCos1.png

... is the basic pattern for integration by parts, and...

http://www.ballooncalculus.org/asy/prod.png

... is the product rule, straight continuous lines differentiating downwards (integrating up) with respect to x, and choosing legs crossed or un-crossed is equivalent to all the labelling with u and v and du and dv.

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