\(\displaystyle \int \frac {x} {(1-4x^2)^\frac {3} {2}} \mathrm{d}x=\)
Let \(\displaystyle u=\frac{1}{\sqrt{1-4x^{2}}}, \;\ \frac{du}{4}=\frac{x}{(1-4x^{2})^{\frac{3}{2}}}dx\)
Now, it turns into a very easy integration.
The second one can be tackled in various ways.
\(\displaystyle \int cos^{4}(x)sin^{2}(2x)dx\)
\(\displaystyle 4\int cos^{4}(x)sin^{2}(x)cos^{2}(x)dx\)
\(\displaystyle 4\int cos^{6}(x)sin^{2}(x)dx\)
\(\displaystyle 4\int cos^{6}(x)(1-cos^{2}(x))dx\)
\(\displaystyle 4\left[\int cos^{6}(x)-\int cos^{8}(x)\right]dx\)
Now, you can use the reduction formula on each one.
\(\displaystyle \int cos^{n}(x)dx=\frac{cos^{n-1}(x)sin(x)}{n}+\frac{n-1}{n}\int cos^{n-2}(x)dx\)
Or, since the powers of sine and cosine are both even and non-negative, you can make repeated use of
\(\displaystyle sin^{2}(x)=\frac{1-cos(2x)}{2}, \;\ cos^{2}(x)=\frac{1+cos(2x)}{2}\)
in \(\displaystyle \int cos^{6}(x)sin^{2}(x)dx\)
