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

1. ## 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...

2. 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.

3. Hello, twostep08!

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

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

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

5. 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.

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