MHB -z.61.W8.6 int sin^2(x)cos^2(x) dx

[karush]

Evaluate $\displaystyle \int\sin^2 \left({x}\right)\cos^2 \left({x}\right)$

$\displaystyle
\sin^2\left({x}\right)
=\frac{1-\cos\left({2x}\right)}{2}$ and $\displaystyle
\cos^2 \left({x}\right)
=\frac{1+\cos\left({2x}\right)}{2}$

So

$\displaystyle
\int\frac{1-\cos\left({2x}\right)}{2}
\cdot
\frac{1+\cos\left({2x}\right)}{2} \ dx
\implies
\frac{1}{4}\int 1-\cos^2 \left({2x} \right) \ dx$

Not sure how to deal with the $\cos^2 \left({2x}\right) $

 

[MarkFL]

Deal with it in the same way you dealt with $\cos^2(x)$ in the original integrand...:)

edit: use a Pythagorean identity first on the new integrand. ;)

 

[suluclac]

$$\frac{1}{4}-\frac{1}{4}\left(\frac{\text{?}}{\text{?}}\right)$$
What Mark is saying here "deal with it in the same way" is applying the simple trig formula for $$\cos^2{2x}$$.
WLOG, $$\cos^2{u}=\frac{1+\cos{2u}}{2}$$.

 

[karush]

Deal with it in the same way you dealt with $\cos^2(x)$ in the original integrand...:)

edit: use a Pythagorean identity first on the new integrand. ;)

$$\frac{1}{4}\int 1-\cos^2 \left({2x} \right) \ dx
\implies
\frac{1}{4}\int \sin^2 \left({2x} \right) \ dx$$

So would the next step be $u=2x$

 

[suluclac]

It could be. If so, what's the next step? What does the sine squared equal to?

 

[Prove It]

Evaluate $\displaystyle \int\sin^2 \left({x}\right)\cos^2 \left({x}\right)$

$\displaystyle
\sin^2\left({x}\right)
=\frac{1-\cos\left({2x}\right)}{2}$ and $\displaystyle
\cos^2 \left({x}\right)
=\frac{1+\cos\left({2x}\right)}{2}$

So

$\displaystyle
\int\frac{1-\cos\left({2x}\right)}{2}
\cdot
\frac{1+\cos\left({2x}\right)}{2} \ dx
\implies
\frac{1}{4}\int 1-\cos^2 \left({2x} \right) \ dx$

Not sure how to deal with the $\cos^2 \left({2x}\right) $

The sine double angle formula may be easier to deal with...

$\displaystyle \begin{align*} \int{ \sin^2{(x)}\cos^2{(x)}\,\mathrm{d}x} &= \int{ \left[ \sin{(x)}\cos{(x)} \right] ^2 \,\mathrm{d}x } \\ &= \int{ \left[ \frac{1}{2}\sin{(2\,x)} \right] ^2 \,\mathrm{d}x } \\ &= \int{ \frac{1}{4} \sin^2{(2\,x)} \,\mathrm{d}x} \\ &= \int{ \frac{1}{4} \,\left\{ \frac{1}{2}\,\left[ 1 - \cos{(4\,x)} \right] \right\} \,\mathrm{d}x} \\ &= \frac{1}{8} \int{ \left[ 1 - \cos{(4\,x)} \right] \,\mathrm{d}x} \end{align*}$

Go from here...

 

[karush]

From the integral table..

$$\displaystyle \int\sin^2 \left({ax}\right) dx
=\frac{x}{2}-\frac{\sin\left({2ax}\right)}{4a}$$

So $a=2$ then

$$\displaystyle
\frac{1 }{4}\left[\frac{x}{2 }+\frac{\sin\left({4x}\right)}{8}\right]
=\frac{1}{8}+\frac{\sin\left({4x}\right)}{32}+C$$