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

**twostep08** · November 12th 2009, 11:07 AM

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...

**adkinsjr** · November 12th 2009, 11:17 AM

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.

**Soroban** · November 12th 2009, 11:20 AM

Hello, twostep08!

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

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

**e^(i*pi)** · November 12th 2009, 11:25 AM

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

**tom@ballooncalculus** · November 12th 2009, 11:41 AM

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...

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

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

... 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.

__________________________________________

Don't integrate - balloontegrate!

Balloon Calculus: Gallery

Balloon Calculus Drawing with LaTeX and Asymptote!

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

LinkBack

Thread Tools

Display

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

Similar Math Help Forum Discussions

Poisson distribution, seems easy enough but I can't figure out what I'm doing wrong

Can't figure out what I'm doing wrong in this logarithm contraction...

two easy questions I can't figure out

Integration By Parts: What am I doing wrong?

Easy question, but cant figure it out

Search Tags