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I have been trying to integrate this problem using trig identities, yet I can never arrive at the right answer.

$$\int \cos^3\left(\frac{x}{3}\right) dx $$

I know that the correct way to approach the problem is with the identity $\cos^2 \theta= 1-\sin^2\theta$, but I'm still having problems.

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you can use substitution for this question: Let $$ u = \frac{x}{3} $$ $$ \frac{du}{dx} = \frac{1}{3} $$ Therefore, $$ 3\cdot du = dx $$.

Now substitute.

$$ \int 3 \cos^3(u)\cdot du$$

Here we can split the integral up like this: $$ 3 \int \cos^2(u) \cos(u) du$$

Using the identity:$$ 1 = \sin^2(x) + \cos^2(x) $$ We can now write our integral as: $$ 3 \int (1-sin^2(u))\cos(u)du $$

Now we have another substitution case: Let $$v = \sin(u)$$ $$dv = \cos(u)du $$

so now our integral will simply be: $$ 3 \int (1 - v^2)dv $$

Now the integral is quite simple and you can proceed. Does that help?

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Use the identity $$\cos^3y=\frac{\cos3y+3\cos y}{4}$$

Disregard my post as OP changed their initial post from $\cos^3(x/3)$ to $\cos(x/3)$.

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You really just want to find $$ \int \cos^3(x)\; dx. $$ (A simple substitution will give you this.)

Now use that $\cos^2(x) = 1 -\sin^2(x)$ so $$ \int \cos^3(x)\; dx = \int \cos^2(x)\cos(x) \; dx = \int (1-\sin^2(x))\cos(x)\;dx. $$ Now use substitution with $u = \sin(x)$ so that $du = \cos(x)dx$. This gives you $$ \int (1-u^2)\;du $$ which I am sure you can do.

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Here's how it works with $\cos^3 u$:

\begin{align} \int \cos^3 u \, du & = \int \cos u (1-\sin^2 u) \, du \\ & = \int \cos u \, du - \int \cos \sin^2 u \, du \\ & = \sin u - \int t^2 \, dt \qquad\qquad \leftarrow t = \sin^2 u, dt = \cos u \, du \\ & = \sin u - \frac{\sin^3 u}{3} \end{align}

(plus a constant).

It should be a simple matter to convert that to work for $\cos^3 \frac{x}{3}$.

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$$\int \cos^3\left(\frac{x}{3}\right) dx $$

Substitute $t=x/3$ and $dt=\frac 1 3dx$

$$=3\int \cos^3\left(t\right) dt $$

Using the reduction formula:

$$=\sin t \cos^2 t+2\int \cos t dt$$

$$=2\sin t+\sin t\cos^2 t+\mathcal C=2\sin (x/3)+\sin (x/3)\cos^2 (x/3)+\mathcal C\\ \color{red}{=\frac 14\left(9\sin\left(\frac x 3\right)+\sin(x)\right)+\mathcal C}$$

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  • $\begingroup$ Can be simplified in some sense by replacing the $\cos^2 t$ by $1-\sin^2 t$ in the last line... $\endgroup$ – Brian Tung Mar 8 '16 at 19:08

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