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

• Nov 12th 2009, 12:07 PM
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, 12:17 PM
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, 12:20 PM
Soroban
Hello, twostep08!

Quote:

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

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

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

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

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

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

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

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

• Nov 12th 2009, 12:25 PM
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

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

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

Edit:

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

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

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

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

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

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

Therefore $-\frac{1}{4}cos(2x)+k = \frac{1}{2}sin^2(x) + C$
• Nov 12th 2009, 12:41 PM
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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