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I am wondering which step has gone wrong. Is it wrong to use $u=\sin x$ ?

$$ \int \cos^3x\ \sin x\;\mathrm{d}x $$ $$=\int\cos^2x \sin x \cos x\;\mathrm{d}x $$ $$=\int(1-\sin^2x)\sin x\;\mathrm{d}(\sin x)$$ $$=\int\sin x\;\mathrm{d}(\sin x)-\int\sin^3x\;\mathrm{d}(\sin x)$$ $$=\frac 12 \sin^2x-\frac14\sin^4x+C$$

Daniel R
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Tiszt
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    it is faster if you take $u=\cos x$ – Nick Oct 18 '16 at 17:22
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    Why are you suspecting it is wrong? – la flaca Oct 18 '16 at 17:23
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    @user56478 I rolled back you edit. Why would you deface the post with a bunch of parantheses when not needed? – Daniel R Oct 18 '16 at 18:41
  • @DanielR The argument of a function such as sine, a logarithm or a summation is the first monomial following it unless an open parenthesis immediately follows the function, in which case the argument is enclosed by that parenthesis and it's counterpart. What the student has written is understandable but wrong, just like how writing $x^2+2x+C=\int 2x+2, dx$ has an inferable meaning but demonstrates incomprehension of the workings of an integral in the truest sense. This post is a prime opportunity to educate in regards to that subject, not to mention the 'slang' usage of an upright $d$. – gen-ℤ ready to perish Oct 18 '16 at 20:20
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    @user56478 I agree that using parentheses is a good way to avoid ambiguity, but in this case the intention is impossible to misinterpret. And I seriously doubt that it is "wrong" as you put it. Please refer me to a mathematical text where your pedantic parenthesizing rules are used; I have never seen one (although I haven't seen that many, I admit that). In this post, my impression (although subjective) was that all those parentheses just cluttered things up and made the post less readable. – Daniel R Oct 18 '16 at 20:57

4 Answers4

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It's fine. You should get an answer that differs from the answer you'd get substituting $u=\cos x$ instead. What's the explanation? The two answers differ by a constant.

But try your method with $\int \cos^2x\sin x\,dx$. Good luck!!

To be a little bit more explicit:

Since $\cos^2 x + \sin^2 x = 1$ you have $$ 1 = (\cos^2 x + \sin^2 x)^2 = \cos^4 x + 2 \sin^2 x \cos^2 x + \sin^4 x $$ Using the identity again on the middle term you have $$ 1 = \cos^4 x - \sin^4 x + 2 \sin^2 x $$ and so the answer you get from the $\sin x$ substitution differ form the $\cos x$ substitution by exactly a constant ($\frac14$ in fact).

Willie Wong
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Ted Shifrin
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no you are not wrong but this can also be done without substitution,

recall your high school formula: $\int x^n\ dx=\frac{x^{n+1}}{n+1}+C$ $$\int \cos^3x\sin x\ dx=-\int (\cos x)^3\ d(\cos x)=-\frac{(\cos x)^4}{4}+C$$

Bhaskara-III
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  • This isn't addressing the original question, however. – Ted Shifrin Oct 18 '16 at 17:33
  • o i am sorry but OP asks high school problem so i give simple method – Bhaskara-III Oct 18 '16 at 17:37
  • I know. He or she knows that this is the usual way of doing the problem, and he's asking if anything is wrong with the different approach. – Ted Shifrin Oct 18 '16 at 17:38
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    You still did substitution, of course. :) – Ted Shifrin Oct 18 '16 at 17:45
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    "...he's asking if anything is wrong with the different approach" , and the answerer he did address the question and answers: "no, you are not wrong"...and then gave a way nicer solution without even needing to specifically substituting (though it would be almost the same, but thriftier this way and it is not formally substitution)> +1 – DonAntonio Oct 18 '16 at 19:58
  • I have $d(\cos x)$ and get the answer. I just can't get the same answer (never will I) from using $d(\sin x)$. Sorry I haven't made it clear in my question. – Tiszt Oct 19 '16 at 14:18
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Your work is correct. You can check your work by differentiating your result: \begin{align*} \frac{d}{dx}\left(\frac 12 \sin^2x-\frac14\sin^4x+C\right)&=\sin x\cos x-\sin^3 x\cos x \\ &=\left(1-\sin^2 x\right)\sin x \cos x \\ &=\sin x\cos^3x \end{align*} This is the original integrand, so your integration was correct.

Aweygan
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  • I have differentiated it and couldn't get the answer. Obviously I have done it wrong... – Tiszt Oct 19 '16 at 13:42
  • Are you talking about the same problem? I did the necessary differentiation in my answer. This shows your integration was correct – Aweygan Oct 19 '16 at 13:54
  • Yes. I know you differentiate it correctly because the answer shown in my question is indeed correct. I just differentiated it incorrectly so I made this post. :( – Tiszt Oct 19 '16 at 13:56
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You can work out such integrals using the complex definition of the trigonometric functions.

With $z=e^{ix}$ (and $dz=ie^{ix}dx$), $\cos x:=\dfrac{z+z^{-1}}2,\sin x=\dfrac{z-z^{-1}}{2i}$ we have

$$ \int \cos^3x\ \sin x\,dx=\int\left(\dfrac{z+z^{-1}}2\right)^3\dfrac{z-z^{-1}}{2i}\frac{dz}{iz}=-\int\frac{z^3+2z-2z^{-3}-z^{-5}}{16}dz\\ =-\frac{\dfrac{z^4+z^{-4}}4+z^2+z^{-2}}{16}=-\frac1{32}\cos4x-\frac18\cos 2x.$$

  • Pardon me, but the font on my computer is glitching on certain words on this page. Could you tell me what the first two words of OP's title are? – user1717828 Oct 19 '16 at 00:09
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    @user1717828 I knew the complex definition of trigonometric functions when I was 13, but I didn't know how to integrate. – Matt Samuel Oct 19 '16 at 00:18
  • @user1717828: High School. –  Oct 19 '16 at 06:34
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    @MattSamuel: the method is powerful in that it turns rational functions of the trigonometric function, even involving multiples of the argument, into plain rational functions. –  Oct 19 '16 at 06:36
  • I don't know what complex definition of the trigonometric functions is... – Tiszt Oct 19 '16 at 13:43
  • @Tiszt: https://en.wikipedia.org/wiki/Trigonometric_functions#Relationship_to_exponential_function_and_complex_numbers. But first you need to know about complex numbers themselves. –  Oct 19 '16 at 13:50
  • @Yves Daoust I only know they are the square root of -1. – Tiszt Oct 19 '16 at 13:52