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I tried to primitive $\sin(x)\cos^3(x)$ step be step, but I got stuck.

Can I use substitution? $[\sin(x)]' = \cos(x)$ And write $\cos^3$ as $\cos^2(x)*\cos(x)$ And also write the function $\sin(x)\cos^2(x)*\cos(x)dx$ as $\sin(x)\cos^2(x)d(\sin(x))$

Brian Fitzpatrick
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Dorians
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    Just let $u=\cos x$. – David Mitra Mar 13 '14 at 09:17
  • @DavidMitra Could you please explain that a bit more? – Dorians Mar 13 '14 at 09:21
  • If $u=\cos x$, then $du=-\sin x ,dx$; so then $\int \sin x\cos^3 x,dx=-\int\cos^3 x (-\sin x),dx=-\int u^3 ,du$. – David Mitra Mar 13 '14 at 09:27
  • What you have done so far is correct; but to compute the antiderivative this way, you would need to write your whole integrand in terms of $\sin x$. The usual mnemonic is to actually make a change of variable; i.e. define $u = \sin x$ and convert the integrand to one in terms of $u$. @David's suggestion is essentially to do the problem in the same way you suggested, but it requires a little less work. –  Mar 13 '14 at 09:29
  • I have never heard the word "primitive" used in this way before. Am I the only one who's puzzled by it?? Actually, I'm pretty sure that "primitive" is not a verb (or, it never used to be, at least). – bubba Mar 13 '14 at 11:02

3 Answers3

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Without substitution:

Suppose you know that $\;\int f(x)dx=F(x)\;$ , and suppose you have a differentiable function $\;g(x)\;$ s.t. $\;f(g(x))\;$ is defined and integrable, then;

$$\int g'(x)\,f(g(x))dx=F(g(x))\;\;\text{(Proof? Apply the Chain Rule to the RHS)}$$

In our case, take

$$\;f(x)=x^3\;\implies \int f(x)dx=\frac14x^4+C\;,\;\;\;g(x)=\cos^3x\;,\;\;\text{so that:}$$

$$\int\sin x\cos^3x\,dx=-\int (-\sin x)\cos^3xdx=-\int(\cos x)'\,\cos^3x\,dx=\frac14\cos^4x+C$$

DonAntonio
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  • You kind of gave a proof of the substitution rule... :) – dani_s Mar 13 '14 at 11:06
  • Yes, I know @dani_s...yet long years of experience with high school and budding undergraduates have taught me that this way, when seen, is simpler and easier to grasp. – DonAntonio Mar 13 '14 at 11:07
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Hint: If you want your " $u$ " to be $\sin x$, it should be fine too. Then write: $$\sin x \cdot \cos ^3 x = \sin x \cdot \cos ^2 x\cdot \cos x = \sin x\cdot (1 - \sin ^2x)\cdot \cos x = (\sin x - \sin ^3 x)\cdot \cos x$$. You can proceed to the next step.

Ant
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DeepSea
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This is almost immediate. Use $y=\cos(x)$, then $dy = -\sin(x)\,dx$, so

$\displaystyle \int dx\,\sin(x)\cos(x) = -\int dy\,y^3 = -\frac{y^4}{4}+C = -\frac{\cos^4(x)}{4}+C$.

jordix
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