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How to find: $$\int^{2\pi}_0 (1+\cos(x))\cos(x)(-\sin^2(x)+\cos(x)+\cos^2(x))~dx$$

I tried multiplying it all out but I just ended up in a real mess and I'm wondering if there is something I'm missing.

Cookie
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Eldat P
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  • Have you tried using $\cos^2(x)-\sin^2(x)=\cos(2x)$ and just integrating? – Moya Jul 25 '15 at 22:45
  • Good call I forgot about that identity. – Eldat P Jul 25 '15 at 22:46
  • Would I need to use parts? – Eldat P Jul 25 '15 at 22:47
  • Definitely, since you're essentially going to end up having integrands which are powers of $\cos$, which can be handled by integration by parts. – Moya Jul 25 '15 at 22:54
  • Could we possibly switch the cosines with complex exponentials and consider the real part of the final result we get or will there be more terms than we desire hanging around, @Moya? – Khallil Jul 25 '15 at 22:56
  • depending on what you are willing to use you could try solving it via complex analytic methods as for example described here: http://mathfaculty.fullerton.edu/mathews/c2003/integralstrigmod.html – jorst Jul 25 '15 at 22:59

3 Answers3

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$\displaystyle\int_0^{2\pi}(\cos x+\cos^2x)(\cos 2x+\cos x)dx=\int_0^{2\pi}\big(\cos x+\frac{1}{2}+\frac{1}{2}\cos 2x\big)\big(\cos 2x+\cos x\big)dx$

$\displaystyle\int_0^{2\pi}\big(\cos x\cos 2x+\cos^{2}x+\frac{1}{2}\cos 2x+\frac{1}{2}\cos x+\frac{1}{2}\cos^{2} 2x+\frac{1}{2}\cos 2x\cos x\big)dx$

$\displaystyle\int_0^{2\pi}\big(\frac{3}{2}\cos x\cos 2x+\frac{1}{2}+\cos 2x+\frac{1}{2}\cos x+\frac{1}{4}+\frac{1}{4}\cos 4x\big)dx$

$\displaystyle\int_0^{2\pi}\big(\frac{3}{4}(\cos 3x+\cos x)+\frac{3}{4}+\cos 2x+\frac{1}{2}\cos x+\frac{1}{4}\cos4x\big)dx=\frac{3}{4}(2\pi)=\frac{3\pi}{2}$

(since $\sin n\pi=0$ for any integer n)

user84413
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You have to linearise the integrand. The simplest way to do is with Euler's formulae. To shorten the computation, set $u=\mathrm e^{\mathrm ix}$. Starting with $\cos x =\dfrac{u+\bar u}2$ and taking into account the relations $$u\bar u=1,\quad u^n+\bar u^n=2\cos nx,$$ we have \begin{align*} (1+\cos x)&\cos x(-\sin^2x+\cos x+\cos^2x)=(1+\cos x)\cos x(\cos x+\cos 2x)\\ &=\Bigl(1+\frac{u+\bar u}2\Bigr)\frac{u+\bar u}2\Bigl(\frac{u+\bar u}2+\frac{u^2+\bar u^2}2\Bigr)\\ &=\frac18(2+u+\bar u)(u+\bar u)\bigl(u+\bar u+u^2+\bar u^2\bigr)\\ &=\frac18\bigr(u^4+\bar u^4+3(u^3+\bar u^3)+4(u^2+\bar u^2)+5(u+\bar u)+6\bigl)\\ &=\frac14(\cos 4x+3\cos 3x+4\cos 2x+5\cos x+3) \end{align*} Consequently the integral from $0$ to $2\pi$ is equal to $$\frac{3\pi}2.$$

Bernard
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try to show that the integrand is equivalent to $$\frac{1}{4} (7 \cos (x)+4 \cos (2 x)+\cos (3 x)+4)$$

Dr. Sonnhard Graubner
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