How can I use integration by parts to solve $\displaystyle \int\sin^2(x) \ dx$?

[karush]

mnt{w.8.4.5} nmh{1000}
$\displaystyle \int\sin^2 \left({x}\right) \ dx
= \frac{x}{2}-\frac{\sin\left({2x }\right)}{4}+C $
As given by a table reference
Integral calculator uses reduction formula to solve this
But this is an exercise following integration by parts so..
$\displaystyle \int\sin^2 \left({x}\right) \ dx
\implies \int\sin\left({x}\right) \sin\left({x}\right) \ dx $
$\displaystyle u=\sin\left({x}\right) $ $dv=\sin\left({x}\right)\ dx$
$du=-\cos\left({x}\right) \ dx $ $v=\cos\left({x}\right)$
$\displaystyle uv-\int v \ du$
$\displaystyle \sin\left({x}\right)\cos\left({x}\right)
+ \int\cos\left({x}\right)\cos\left({x}\right) \ dx $
Got stuck here

 

[Prove It]

Whitman 8.4.5
$\displaystyle \int\sin^2 \left({x}\right) \ dx
= \frac{x}{2}-\frac{\sin\left({2x }\right)}{4}+C $
As given by a table reference
Integral calculator uses reduction formula to solve this
But this is an exercise following integration by parts so..
$\displaystyle \int\sin^2 \left({x}\right) \ dx
\implies \int\sin\left({x}\right) \sin\left({x}\right) \ dx $
$\displaystyle u=\sin\left({x}\right) $ $dv=\sin\left({x}\right)\ dx$
$du=-\cos\left({x}\right) \ dx $ $v=\cos\left({x}\right)$
$\displaystyle uv-\int v \ du$
$\displaystyle \sin\left({x}\right)\cos\left({x}\right)
+ \int\cos\left({x}\right)\cos\left({x}\right) \ dx $
Got stuck here

I don't know why you always stop at this step. It's pretty obvious that you still have an integration to do, and since you were using integration by parts for sin^2(x), wouldn't it make sense that you would have to use integration by parts for cos^2(x) as well?

 

[soroban]

Whitman 8.4.5
$\displaystyle \int\sin^2 \left({x}\right) \ dx
= \frac{x}{2}-\frac{\sin\left({2x }\right)}{4}+C $
As given by a table reference
Integral calculator uses reduction formula to solve this
But this is an exercise following integration by parts so..
$\displaystyle \int\sin^2 \left({x}\right) \ dx
\implies \int\sin\left({x}\right) \sin\left({x}\right) \ dx $
$\displaystyle u=\sin\left({x}\right) $ $dv=\sin\left({x}\right)\ dx$
$du=-\cos\left({x}\right) \ dx $ $v=\cos\left({x}\right)$
$\displaystyle uv-\int v \ du$
$\displaystyle \sin\left({x}\right)\cos\left({x}\right)
+ \int\cos\left({x}\right)\cos\left({x}\right) \ dx $
Got stuck here

Your work is a little off . . .

$$I \;=\;\int\sin^2x\,dx \;=\;\int\underbrace{\sin x}_u\,\underbrace{\sin x\,dx}_{dv}$$

$$\begin{array}{ccc}u \:=\:\sin x && dv \:=\:\sin x\,dx \\ du \:=\:\cos x\,dx && v \:=\:-\cos x \end{array}$$

$$I \;=\;\underbrace{(\sin x)}_u \underbrace{(-\cos x)}_v - \int\underbrace{(-\cos x)} _v\underbrace{(\cos x\,dx)}_{du} \;=\; -\sin x\cos x + \int\cos^2x\,dx$$

$$I \;=\; -\sin x\cos x + \int(1 -\sin^2x\,dx)\,dx \;=\;-\sin x\cos x + \int dx - \underbrace{\int\sin^2x\,dx}_{\text{This is }I}$$

$$I \;=\;-\sin\cos x + x - I \quad\Rightarrow\quad 2I \;=\;x -\sin x\cos x +C$$

$$I \;=\;\frac{1}{2}(x - \sin x\cos x) + C$$

 

[Greg]

\(\displaystyle \int\sin^2(x)\,dx\)

\(\displaystyle dv=1,v=x\)

\(\displaystyle u=\sin^2(x),du=\sin(2x)\,dx\)

\(\displaystyle \int\sin^2(x)\,dx=x\sin^2(x)-\int x\sin(2x)\,dx\)

IBP (again) with $u=x,dv=\sin(2x)$:

\(\displaystyle \int\sin^2(x)\,dx=x\sin^2(x)-\left(-x\dfrac{\cos(2x)}{2}+\dfrac12\int\cos(2x)\,dx\right)\)

\(\displaystyle =x\sin^2(x)+x\dfrac{\cos(2x)}{2}-\dfrac{\sin(2x)}{4}+C=\dfrac x2-\dfrac{\sin(2x)}{4}+C\)

 

[karush]

Thanks everyone sorry my latex didn't render to good in the preview it was fine. I'm too used to just looking things up in the tables not knowing how it was derived the book just had examples of odd powers.