solving $\int \cos^2x \sin2x dx$ 
$$\int \cos^2x \sin2x dx$$

$$\int \cos^2x \sin2x \, dx=\int \left(\frac{1}{2} +\frac{\cos2x}{2} \right) \sin2x \, dx$$
$u=\sin2x$
$du=2\cos2x\,dx$
$$\int \left(\frac{1}{2} +\frac{du}{4}\right)u \, du$$
Is the last step is ok?
 A: Since $\sin 2x=2\sin x\cos x$ we have
$$\int\cos^2 x\sin 2x\,dx=\int 2\cos^3 x\sin x\,dx=-\frac{1}{2}\cos^4 x+C$$
A: The correct substition is $u = \cos(2x)$, $du = -2\sin(2x)\,dx$, so you 
get 
$$\int(1/2 + u/2)(du/(-2)).$$
A: Using the substitution you proposed $u\leadsto\sin2x$ one would get 
$$\begin{align}
\int\cos^2x\sin2x\,\mathrm dx&=\int\left(\dfrac12+\dfrac{\cos 2x}2\right)\sin 2x\,\mathrm dx\\
&=\int\dfrac{\sin 2x}2\,\mathrm dx+\int\dfrac{\sin 2x}{4}\underbrace{{2\cos 2x}\,\mathrm dx}_{\displaystyle\mathrm du}\\
&=\int\dfrac{\sin 2x}{2}\,\mathrm dx+\int\dfrac{u}{4}\,\mathrm du,
\end{align}$$ which is the correct form.
A: You might have set $u=\cos 2x$, hence $ \mathrm d\mkern1mu u = -2\sin 2x\mathrm d\mkern1mu x$, whence
\begin{align*}\int\cos^2 x\sin 2x\,\mathrm d\mkern1mu x&=\int-\frac14(1+u)\,\mathrm d\mkern1mu u = -\frac14\Bigl(u+\frac{u^2}2\Bigr)\\
&=-\frac14\cos 2x-\frac18\frac{1+\cos4 x}2=-\frac14\cos 2x-\frac1{16}\cos 4x+C.
\end{align*}
A: General strategy in such cases (trigonometric functions, multiples of some angles) is to express everything in terms of $\sin x$ and $\cos x$, and take it from there.
A: Alternatively, write $\sin2x=2\sin x\cos x$ and integrate $2\cos^3x \sin x$ and get $-\frac 12\cos^4 x+c$
A: $$\int  cos^{ 2 }xsin2xdx=2\int { \cos ^{ 3 }{ x\sin { xdx= }  } -2\int { \cos ^{ 3 }{ x } d\cos { x }  } =-\frac { \cos ^{ 4 }{ x }  }{ 2 }  } +C$$
A: $$\int  \cos^{ 2 }x\sin2xdx\\=2\int { \cos ^{ 3 }{ x\sin { xdx= }  } -2\int { \cos ^{ 3 }{ x } d\cos { x }  } \\=\boxed{\color{blue}{-\frac {\cos ^{ 4 } x   }{ 2 }  +C}}}$$
