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So I'm trying to figure out how the reduction formula can help me solve integrands like $\int \cos^3(x)\,dx$. I've figured that if I'd use integration by parts, I would get to the following: $$\int u\,dv = uv - \int v\,du$$ $$u = \cos^m(x)$$ $$du = -m\sin(x)\cos^{m-1}\,dx$$ $$dv = dx$$ $$v = x$$

$$\int \cos^m(x)\,dx = x\cos^m + \int x \cdot m\cdot \sin(x)\cos^{m-1}\, dx$$

Which 1. doesn't seem to really help me and 2. doesn't seem to be true.

Would really appreciate some help, thanks in advance!

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    $\begingroup$ Try replacing a $\cos^2(x)$ with $(1-\sin^2(x))$. $\endgroup$ – Arturo Magidin Mar 3 at 19:22
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    $\begingroup$ tutorial.math.lamar.edu/classes/calcII/IntegralsWithTrig.aspx is a really good resource for tools for these integrals $\endgroup$ – Alan Mar 3 at 19:25
  • $\begingroup$ It's often useful to rewrite the expression as sum of $\cos kx$ and $\sin kx$. For example, you can rewrite $\cos\alpha\cos\beta$ as $\frac{1}{2}(\cos (\alpha+\beta)+\cos(\alpha-\beta))$. $\endgroup$ – richrow Mar 3 at 19:37
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$$\int \cos^3(x)\,dx= \int (1- \sin^2(x))\cos(x)\,dx$$

and let $u= \sin(x)$.

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This is a standard exercise. $$\int \cos^3{x}\,dx =\int \cos^2{x} \cos{x}\,dx =\int (1-\sin^{2}{x}) \cos{x}\,dx=$$ $$=\int \cos{x}\,dx-\int \sin^{2}{x} \cos{x}\,dx=\sin{x}-\frac{\sin^{3}{x}}{3}+C. $$

This method is useful in finding $\int \cos^m{x}\,dx$ and $\int \sin^m{x}\,dx$ whenever $m$ is an odd integer.

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$$\cos(3x)=\cos(2x)\cos(x)-\sin(2x)\sin(x)$$ $$=(2\cos^2(x)-1)\cos(x)-2\cos(x)\sin^2(x)$$

$$=4\cos^3(x)-3\cos(x)$$

So

$$4\cos^3(x)=\cos(3x)+3\cos(x)$$ and

$$4\int \cos^3(x)dx=\frac{\sin(3x)}{3}+3\sin(x)+C$$

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It is simpler to use the standard method: linearising $\cos^3x=\frac14\bigl(\cos3x+3\cos x)$, which yields instantly $$\int\cos^3x\,\mathrm dx=\frac14\biggl(\int\cos 3x\,\mathrm dx+3\int\cos x\,\mathrm dx\biggr).$$

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Integrating by parts, take

$$u=\cos^2x\implies \mathrm du=-2\cos x\sin x\,\mathrm dx\\ \mathrm dv=\cos x\,\mathrm dx\implies v=\sin x$$

(Picking $\mathrm dv=\mathrm dx$, as you've done, indeed isn't of much help.)

Then

$$\begin{align} \int\cos^3x\,\mathrm dx&=uv-\int v\,\mathrm du\\[1ex] &=\cos^2x\sin x+2\int\cos x\sin^2x\,\mathrm dx \end{align}$$

In the remaining integral, substitute $t=\sin x$ and $\mathrm dt=\cos x\,\mathrm dx$ to get

$$\begin{align} \int\cos^3x\,\mathrm dx&=\cos^2x\sin x+2\int t^2\,\mathrm dt\\[1ex] &=\cos^2x\sin x+\frac23t^3+C\\[1ex] &=\cos^2x\sin x+\frac23\sin^3x+C \end{align}$$

Differentiate to confirm:

$$\frac{\mathrm d}{\mathrm dx}\left[\cos^2x\sin x+\frac23\sin^3x+C\right]=(-2\cos x\sin^2x+\cos^3x)+2\sin^2x\cos x\stackrel{\checkmark}{=}\cos^3x$$

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