Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

Thanks in advance,

share|improve this question
1  
$(1+\cos\;x)^3=\frac14(10+15\cos\;x+6\cos\;2x+\cos\;3x)$ –  J. M. Dec 23 '10 at 15:10
3  
Or integrate by parts, with sufficient persistence. –  Mariano Suárez-Alvarez Dec 23 '10 at 15:21

4 Answers 4

up vote 8 down vote accepted

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.

share|improve this answer
    
For $\cos^3x$ we could also use the identity $\cos 3x = 4 \cos ^3x - 3 \cos x$ and it would be simple :) –  Quixotic 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.

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

share|improve this answer

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*}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.