Integrate the expression

Question

There is a task given, but I am not sure if it is right, because I don't have a clue on how to solve it. Even tried to put it on wolfram to get any result, but no chance. Would appreciate your help: $$\int \sin^42\cos^22x\ \mathrm dx$$

Answer 1

Assuming that the integral is:

$$\int\sin^4 (2\color{red}x)\cos^2(2x)dx$$

Hint:

Substitute $t=2x$ and $dt=2dx$

$$=\frac 1 2\int \sin^4 t \cos^2 t dt$$

Write as:

$$=\frac 1 2\int \sin^4 t(1-\sin^2 t)dt$$

$$=-\frac 1 2\int \sin^6 t dt +\frac 1 2\int\sin^4 t dt$$

Using the reduction formula:

$$=\frac{1}{12}\sin^5 t\cos t+\frac{1}{12}\sin^4 t dt$$

Reduction formula again:

$$-\frac{1}{48}\sin^3 t\cos t+\frac{1}{12}\sin^5t\cos t+\frac{1}{16}\int \sin^2 t dt$$

Now use the fact that $\color{green}{\sin^2 t=\frac 1 2-\frac 1 2 \cos(2t)}$ and now it is easy to finish.

Answer 2

Interpreting as:

$$\int (\sin^4 2)(\cos^2 2x)dx$$

Than $\sin^4 2=a \approx 0.6836343..$ is a constant and the integral becomes

$$\int a\cos^2 2xdx$$

that can be integrated noting that $$\cos^2 2x=\frac{1}{2}(\cos 4x-1)$$ and substituting $4x=t$