# How can I find $\int{(1 + \cos x)^3\mathrm dx}$?

My question is ; How can I solve the following integral question?

$$\int{(1 + \cos x)^3\mathrm dx}$$

Thanks in advance,

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$(1+\cos\;x)^3=\frac14(10+15\cos\;x+6\cos\;2x+\cos\;3x)$ –  Guess who it is. Dec 23 '10 at 15:10
Or integrate by parts, with sufficient persistence. –  Mariano Suárez-Alvarez Dec 23 '10 at 15:21

## 4 Answers

If you don't know how to rewrite $\cos^n(x)$ in terms of $\cos(nx)$, here is an alternate solution.

Expand $(1+\cos(x))^3 = 1 + 3\cos(x) + 3\cos^2(x) + \cos^3(x)$. Now, you know how to integrate the first two terms.

The third term can be integrated by using the half-angle formula $$\cos^2(x) = \frac{1+\cos(2x)}{2},$$ so that $$\int 3\cos^2(x)\,dx = 3\left(\frac{x}{2} + \frac{\sin(2x)}{4}\right)$$

The last term can be integrated using the pythagorean identity $\cos^2(x) = 1 - \sin^2(x),$ so $$\cos^3(x) = \cos(x) - \cos(x)\sin^2(x).$$ And now $\int \cos(x)\,dx = \sin x$ and $\int \cos(x)\sin^2(x)\,dx = \frac{\sin^3(x)}{3}$, so $$\int \cos^3(x)\,dx = \sin x - \frac{\sin^3(x)}{3}.$$

Now just put the pieces together.

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For $\cos^3x$ we could also use the identity $\cos 3x = 4 \cos ^3x - 3 \cos x$ and it would be simple :) –  VelvetThunder Dec 23 '10 at 21:10

You can also use complex numbers to make it mechanical.

Write $\displaystyle \cos x = \dfrac{e^{ix} + e^{-ix}}{2}$ and just expand out.

This method makes finding integrals of the form

$$\int P(\sin x, \cos x) \ \text{dx}$$

(where $P(x,y)$ is a polynomial in the variables $x,y$) purely mechanical, without having to do any clever algebra etc.

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Totally agree with the method, but the idiots in education system want some type of specific way of doing things, as swift as this method is, at some point a bozo will come up with "But that is not how we do it in this course" comment and tries not to give the marks due to a correct answer. So many examples of this type of attitude in maths teaching makes one wonder if mathematical truths are universal or localized to specific courses. –  Arjang Dec 24 '10 at 22:31

Expand $(1+\cos(x))^3$. And rewrite $\cos^n(x)$ in terms of $\cos(nx)$ and other similar terms.

$(1+\cos(x))^3 = 1 + 3 \cos(x) + 3 \cos^2(x) + \cos^3(x) = \frac{\cos(3x)+6 \cos(2x)+15 \cos(x)+10}{4}$.

Now, integrate term by term to get the desired answer.

$\int \cos(nx) = \frac{\sin(nx)}{n}$.

Hence, $\int (1+\cos(x))^3 = \frac{\frac{\sin(3x)}{3}+6\frac{\sin(2x)}{2}+15 \sin(x) + 10x}{4} + c = \frac{\sin(3x)}{12} + \frac{3\sin(2x)}{4} + \frac{15 \sin(x)}{4} + \frac{5x}{2} + c$.

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HINT:

You can use the Half-angle formula. $$\cos{2x} = 2\cos^{2}{x}-1$$ and so you have $$1+\cos{x}= 1 + 2\cos^{2}\frac{x}{2} -1 =2\cos^{2}\frac{x}{2}$$ So the required integral is

\begin{align*} \int (1+\cos{x})^{3} \rm{dx} &= \int 8 \cdot \cos^{6}\Bigl(\frac{x}{2}\Bigr) \ \rm{dx} \\ &= 8 \int \cos^{6}\Bigl(\frac{x}{2}\Bigr) \ \rm{dx} \end{align*}

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