The value of the integral _0^/2(60(6x))/( x),dx is equal to________.

Question

The value of the integral $\displaystyle\int_0^{\pi/2}\frac{60\sin(6x)}{\sin x}\,dx$ is equal to\_\_\_\_\_\_\_\_.

Answer: 104

Accepted Answer

Using $\sin6x = 2\sin3x\cos3x$ and $\sin3x=\sin x(3-4\sin^2x)$: $\dfrac{\sin6x}{\sin x} = 2\cos3x(3-4\sin^2x)$, and with $\cos3x=4\cos^3x-3\cos x=(4\cos^2x-3)\cos x$... following the source's route:

$$I = 60\cdot2\int_0^{\pi/2}(3-4\sin^2x)(4\cos^2x-3)\cos x\,dx = 120\int_0^{\pi/2}(3-4\sin^2x)(1-4\sin^2x)\cos x\,dx$$

With $t=\sin x$:

$$I = 120\int_0^1(3-16t^2+16t^4)\,dt = 120\left[3t-\frac{16t^3}3+\frac{16t^5}5\right]_0^1 = 104$$

Solutions · 1