# Prove that integral equals zero

Prove that an integral

$$\int_0^{\pi/2} \cos^{10}x\cdot \cos(12x)\,\text{d}x=0$$

I'm sorry but I am completely lost. As far as I know, an integral is $0$ when:

1) $f(x)=0$ in every $x$ in $[a,b]$, which doesn't apply here.

2) If $f(x)$ is odd and upper and lower bounds are symmetric.

3) Bounds are equal

None of which seem to be the case. I figured out that the $f(x)$ is even but where do I go from here. I tried to integrate this by parts but it got real nasty, there should be a simpler way right?

• $\int_a^b f(x) dx = \int_a^b f (a+b-x) dx$ Commented Nov 14, 2017 at 10:56

In a more general case than the three examples you mention, an integral is $0$ whenever the positive (above the $x$-axis) and negative (below the $x$-axis) areas in the interval $[a,b]$ have exactly the same absolute value in total so that they cancel out due to the signs. This integral however doesn't have an obvious symmetry that allows to quickly showcase it equaling $0$, so - instead - it can just be computed normally with the help of the cosine addition formula: \begin{align} \int\cos^{10}(x)\cos(12x)\text{d}x&=\int\cos^{10}(x)\left(\cos(11x)\cos(x)-\sin(11x)\sin(x)\right)\text{d}x\\ &=\int\cos^{11}(x)\cos(11x)\text{d}x-\int\cos^{10}x\sin(11x)\sin(x)\text{d}x\\ \text{Now use IBP for the }&\text{left integral with $u'=\cos(11x)$ and $v=\cos^{11}(x)$}\\ &=\frac{1}{11}\sin(11x)\cos^{11}(x)-\int\frac{1}{11}\sin(11x)\cdot11\cos^{10}(x)(-\sin(x))\text{d}x\\&\hspace{0.4cm}-\int\cos^{10}x\sin(11x)\sin(x)\text{d}x\\ &=\frac{1}{11}\sin(11x)\cos^{11}(x)+\int\cos^{10}x\sin(11x)\sin(x)\text{d}x\\ &\hspace{0.4cm}-\int\cos^{10}x\sin(11x)\sin(x)\text{d}x\\ &=\frac{1}{11}\sin(11x)\cos^{11}(x)+C \end{align} From this it is clear that the integral in question equals $0$.
$$2^{10}\cos^{10}x=(e^{ix}+e^{-ix})^{10}=2\left(\cos10x+\binom{10}1\cos8x+\binom{10}2\cos6x+\binom{10}3\cos4x+\binom{10}4\cos2x+\binom{10}5\right)$$
For example: $$\int_0^{\pi/2}2\cos8x\cos12x\ dx=\int_0^{\pi/2}(\cos4x+\cos20x)dx=?$$
I have a simple to get the answer. In fact, under $u=\frac{\pi}{2}-x$ \begin{eqnarray} I&=&\int_0^{\pi/2} \cos^{10}x\cos(12x)\,\text{d}x\\ &=&-\int_{0}^{\pi/2} \cos^{10}(\frac{\pi}{2}-u)\ \cos(12(\frac{\pi}{2}-u))\,\text{d}u\\ &=&-\int_{0}^{\pi/2} \sin^{10}u\cos(4\pi-12u)\,\text{d}u\\ &=&-\int_{0}^{\pi/2} \sin^{10}u \cos(12u)\,\text{d}u \end{eqnarray} and hence $$2I=\int_0^{\pi/2} [\cos^{10}x-\sin^{10}x]\cos(12x)\,\text{d}x.$$ In fact, under $u=\frac{\pi}{2}-x$, $$2I=\int_0^{\pi/2} [\sin^{10}x-\cos^{10}x]\cos(12x)\,\text{d}x=-2I$$ and hence $$I=0.$$