$$\int \cos^2x \sin2x dx$$
$$\int \cos^2x \sin2x \, dx=\int \left(\frac{1}{2} +\frac{\cos2x}{2} \right) \sin2x \, dx$$
$u=\sin2x$
$du=2\cos2x\,dx$
$$\int \left(\frac{1}{2} +\frac{du}{4}\right)u \, du$$
Is the last step is ok?
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3No. What is $\int \frac12u$ without any $\mathrm d\mkern1mu u$? – Bernard Jan 03 '16 at 19:28
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@Bernard $du$ must multiply all the integrand? – gbox Jan 03 '16 at 19:31
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2Absolutely. Otherwise, this notation is meaningless. – Bernard Jan 03 '16 at 19:41
8 Answers
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Since $\sin 2x=2\sin x\cos x$ we have $$\int\cos^2 x\sin 2x\,dx=\int 2\cos^3 x\sin x\,dx=-\frac{1}{2}\cos^4 x+C$$
Ángel Mario Gallegos
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The correct substition is $u = \cos(2x)$, $du = -2\sin(2x)\,dx$, so you get $$\int(1/2 + u/2)(du/(-2)).$$
ncmathsadist
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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.
Workaholic
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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*}
Bernard
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1@gbox Use the fact that $u=\cos2x=2\cos^2x-1$, hence $\cos^2x=\dfrac{u+1}2$. – Workaholic Jan 03 '16 at 19:46
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2@gbox: I started from where you arrived just before making your substitution. – Bernard Jan 03 '16 at 19:48
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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.
vonbrand
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Alternatively, write $\sin2x=2\sin x\cos x$ and integrate $2\cos^3x \sin x$ and get $-\frac 12\cos^4 x+c$
David Quinn
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$$\int cos^{ 2 }xsin2xdx=2\int { \cos ^{ 3 }{ x\sin { xdx= } } -2\int { \cos ^{ 3 }{ x } d\cos { x } } =-\frac { \cos ^{ 4 }{ x } }{ 2 } } +C$$
haqnatural
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$$\int \cos^{ 2 }x\sin2xdx\\=2\int { \cos ^{ 3 }{ x\sin { xdx= } } -2\int { \cos ^{ 3 }{ x } d\cos { x } } \\=\boxed{\color{blue}{-\frac {\cos ^{ 4 } x }{ 2 } +C}}}$$
3SAT
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