# 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?

• No. What is $\int \frac12u$ without any $\mathrm d\mkern1mu u$? – Bernard Jan 3 '16 at 19:28
• @Bernard $du$ must multiply all the integrand? – gbox Jan 3 '16 at 19:31
• Absolutely. Otherwise, this notation is meaningless. – Bernard Jan 3 '16 at 19:41

Since $\sin 2x=2\sin x\cos x$ we have $$\int\cos^2 x\sin 2x\,dx=\int 2\cos^3 x\sin x\,dx=-\frac{1}{2}\cos^4 x+C$$

• sweet and slick – ncmathsadist Jan 3 '16 at 20:13

The correct substition is $u = \cos(2x)$, $du = -2\sin(2x)\,dx$, so you get $$\int(1/2 + u/2)(du/(-2)).$$

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.

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*}

• how did you substitute $$\cos^2x$$? – gbox Jan 3 '16 at 19:42
• @gbox Use the fact that $u=\cos2x=2\cos^2x-1$, hence $\cos^2x=\dfrac{u+1}2$. – Workaholic Jan 3 '16 at 19:46
• @gbox: I started from where you arrived just before making your substitution. – Bernard Jan 3 '16 at 19:48

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.

Alternatively, write $\sin2x=2\sin x\cos x$ and integrate $2\cos^3x \sin x$ and get $-\frac 12\cos^4 x+c$

$$\int cos^{ 2 }xsin2xdx=2\int { \cos ^{ 3 }{ x\sin { xdx= } } -2\int { \cos ^{ 3 }{ x } d\cos { x } } =-\frac { \cos ^{ 4 }{ x } }{ 2 } } +C$$

$$\int \cos^{ 2 }x\sin2xdx\\=2\int { \cos ^{ 3 }{ x\sin { xdx= } } -2\int { \cos ^{ 3 }{ x } d\cos { x } } \\=\boxed{\color{blue}{-\frac {\cos ^{ 4 } x }{ 2 } +C}}}$$