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I'm trying to solve $\int{\sin^3(x)\cos^2(x)}dx$.

I got $-\frac{1}{2}\cos(x)+C$, but the memo says $\frac{1}{5}\cos^5(x)-\frac{1}{3}\cos^3(x)+C$

This is my working:

enter image description here

Your help is appreciated!


marked as duplicate by Jyrki Lahtonen Jun 15 '16 at 13:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Hint: Substitute $u=\cos x$. $\endgroup$ – Quang Hoang Aug 19 '14 at 7:28
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    $\begingroup$ How do you go from the second line to the third line? $\endgroup$ – user37238 Aug 19 '14 at 7:28
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    $\begingroup$ For accuracy in these type of calculations, its important to be able to realize when an answer you got is complete nonsense (because everyone makes mistakes in calculations sometimes). The answer you got is very easy to differentiate. It should be clear to you why its incorrect. $\endgroup$ – PVAL-inactive Aug 19 '14 at 7:33
  • $\begingroup$ $1 - \sin^2x - \cos^2x + \sin^2x \cdot \cos^2x \\= 1 -(\sin^2 x + \cos^2 x) + \sin^2 x \cdot\cos^2 x \\= 0 + \sin^2x\cdot\cos^2x$ $\endgroup$ – Nick Aug 19 '14 at 8:57
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    $\begingroup$ This post also says something about integrals of the form $\int \sin^n x \cos^m x \,\mathrm{d}x$: math.stackexchange.com/questions/29980/… $\endgroup$ – Martin Sleziak Aug 19 '14 at 11:34

Rewrite as

$$\int(1-\cos^2 x) \cos^2 x \sin x \ dx$$

and let $t=\cos x\Rightarrow dt = -\sin x \ dx$.


Hint :

It will be easy if you didn't write $\cos^2x$ as $1-\sin^2x$. Write the integrand as $\sin x(1-\cos^2x)\cos^2x$ then set $t=\cos x$.


I was challenged to do this question without using either substitution or even $\sin^2 x + \cos^2x = 1$,

$\sin^3 x \cdot\cos^2 x \\ = \sin x \cdot \sin^2 x \cdot \cos^2 x\\ = \sin x \cdot (\frac{1}{2}\cdot 2\sin x\cdot\cos x)^2\\ = \sin x \cdot (\frac{1}{2} \sin 2x)^2\\ = \frac{1}{4}\cdot (\sin x \cdot \sin 2x) \cdot \sin 2x\\ = \frac{1}{4} \cdot \frac{1}{2} (\cos x - \cos 3x)\cdot \sin 2x\\ = \frac{1}{8}\cdot (\sin2x\cdot \cos x - \sin2x\cdot \cos3x)\\ =\frac{1}{8}\cdot[\frac{1}{2}(\sin x + \sin 3x) - \frac{1}{2}(\sin(-x) + \sin5x) ]\\ =\frac{1}{8}\cdot\frac{1}{2} (\sin x + \sin 3x + \sin x - \sin 5x)\\ =\frac{1}{16}(2\sin x + \sin3x - \sin 5x)\\$

$$\therefore \int{(\sin^3x\cdot\cos^2x)}\cdot dx \\ = \frac{1}{16}\int{(2\sin x + \sin3x - \sin 5x)}\cdot dx\\ = \frac{(-\cos x)}{8} + \frac{(-\cos3x)}{3\times 16} - \frac{(-\cos5x)}{5\times 16} + C\\ = \frac{\cos5x}{80} - \frac{\cos3x}{48} - \frac{\cos x}{8} + C$$

$\dots$which isn't the answer in the memo but that doesn't mean it isn't right.

Now, if there's any math teacher out there who won't accept my answer on a written examination, please speak or forever hold your peace.

Edit: This video could useful for general questions of this type

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    $\begingroup$ WA says that both answers are equal: tinyurl.com/mqox8rv $\endgroup$ – Martin Sleziak Aug 19 '14 at 10:51
  • $\begingroup$ @MartinSleziak: Ofcourse they are, I merely left the proof of that as an exercise to the reader. Also, I'd like to take this opportunity to say that I haven't violated community spirit by giving a direct answer. I simply illustrated that in antiderivitives, answers can vary depending on the simplifications used. $\endgroup$ – Nick Aug 19 '14 at 10:59
  • $\begingroup$ How did you get $\sin x\cdot \sin 2x = \frac{1}{2}(\cos x - \cos 3x)$? and $\frac{1}{8}\cdot (\sin2x\cdot \cos x - \sin2x\cdot \cos3x) =\frac{1}{8}\cdot[\frac{1}{2}(\sin x + \sin 3x) - \frac{1}{2}(\sin(-x) + \sin5x)$? $\endgroup$ – ahorn Aug 19 '14 at 13:19

Just to comment on your attempted solution (from the picture in your post).

You wrote: $$ \int \sin^3 x \cos^2x\,\mathrm{d}x = \\ \int \sin x (1-\cos^2x)(1-\sin^2x)\,\mathrm{d}x=\\ \int \sin x (1-\cos^2x-\sin^2x+\cos^2x\sin^2x)\,\mathrm{d}x\overset{*}=\\ \int \sin x (1-\cos^2x\sin^2x)\,\mathrm{d}x=\\ \int \sin x \,\mathrm{d}x - \int \sin^3 x \cos^2x\,\mathrm{d}x $$ $2I=\int \sin x \,\mathrm{d}x$ $I=-\frac12\cos x+C$

I think the mistake is in the step marked by $(*)$, where you equated these two things $$1-\cos^2x-\sin^2x+cos^2x\sin^2x \overset{?}= 1-\cos^2x\sin^2x$$ If you plug in $x=0$, you will see that they are not equal. (LHS is equal to 0 for $x=0$ and the RHS is equal to 1.)

This has already been mentioned in comments.


This can be rewritten as $ \int \sin^2(x) \cos^2(x) \sin(x) dx $

using the property that $ \sin^2(x) = 1 - \cos^2(x) $, we can rewrite the original integral as

$ \int (1 - \cos^2(x)) \cos^2(x) \sin(x)dx $

substitute $u = \cos(x)$ gets us

$ -\int (1 - u^2)(u^2)du $

= $ -\int u^2 - u^4 du$

now we have an easy integrand to work with:

$ -\int u^2 - u^4 du = -(\frac{u^3}{3} - \frac{u^5}{5} + C) $

= $\frac{u^5}{5} - \frac{u^3}{3} + C$

substitute $ u = \cos(x) $ to obtain the final answer

$\frac{\cos^5(x)}{5} - \frac{cos^3(x)}{3} + C$


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