# Question #5ea5f

Jun 21, 2016

I found: $\frac{1}{2} \left[x - \sin \left(x\right) \cos \left(x\right)\right] + c$

#### Explanation:

Try this:

Jun 21, 2016

Alternatively, you could make use of trig identities to find the same result: $\int {\sin}^{2} x \mathrm{dx} = \frac{1}{2} \left(x - \sin x \cos x\right) + C$

#### Explanation:

In addition to Gio's method, there is another way of doing this integral, using trig identities. (If you don't like trig or math in general, I wouldn't blame you for disregarding this answer - but sometimes the use of trig is unavoidable in problems).

The identity we will be using is: ${\sin}^{2} x = \frac{1}{2} \left(1 - \cos 2 x\right)$.

We can therefore rewrite the integral like so:
$\int \frac{1}{2} \left(1 - \cos 2 x\right) \mathrm{dx}$
$= \frac{1}{2} \int 1 - \cos 2 x$

Using the sum rule we get:
$\frac{1}{2} \left(\int 1 \mathrm{dx} - \int \cos 2 x \mathrm{dx}\right)$

The first integral simply evaluates to $x$. The second integral is a little more challenging. We know that the integral of $\cos x$ is $\sin x$ (because $\frac{d}{\mathrm{dx}} \sin x = \cos x$), but what about $\cos 2 x$? We'll need to adjust for the chain rule by multiplying by $\frac{1}{2}$, so as to balance the $2 x$:
$\frac{d}{\mathrm{dx}} \frac{1}{2} \sin 2 x = 2 \cdot \frac{1}{2} \cos 2 x = \cos 2 x$

So $\int \cos 2 x \mathrm{dx} = \frac{1}{2} \sin 2 x + C$ (don't forget the integration constant!) Using that info, plus the fact that $\int 1 \mathrm{dx} = x + C$, we have:
$\frac{1}{2} \left(\textcolor{red}{\int 1 \mathrm{dx}} - \textcolor{b l u e}{\int \cos 2 x \mathrm{dx}}\right) = \frac{1}{2} \left(\textcolor{red}{x} - \textcolor{b l u e}{\frac{1}{2} \sin 2 x}\right) + C$

Use the identity $\sin 2 x = 2 \sin x \cos x$, we find:
$\frac{1}{2} \left(x - \frac{1}{2} \sin 2 x\right) + C = \frac{1}{2} \left(x - \frac{1}{2} \left(2 \sin x \cos x\right)\right) + C$
$= \frac{1}{2} \left(x - \sin x \cos x\right) + C$

And that is the answer Gio found using the integration by parts method.