Sine Integration

Calculus Level 5

$\large\displaystyle\int_0^\pi\sin^4(x+\sin(3x))\, dx$

If the integral above equals to $\dfrac{a\pi}b$ , where $a$ and $b$ are coprime positive integers, find $a+b$ .

1 solution

Jack Lam
Jul 7, 2016

Identity 1:

$8\sin^4{\theta} \equiv \cos{4\theta} - 4\cos{2\theta} + 3$

Proof:

$2i\sin{\theta} \equiv e^{ix} - e^{-ix}$

$(2i\sin{\theta})^4 \equiv e^{4ix} + e^{-4ix} - 4(e^{2ix} + e^{-2ix}) + 6$

$16\sin^4{\theta} \equiv 2\cos{4\theta} - 8\cos{2\theta} + 6$

$8\sin^4{\theta} \equiv \cos{4\theta} - 4\cos{2\theta} + 3$

Identity 2:

$\cos{\theta} - \cos{(\theta + \frac{\pi}{3})} - \cos{(\theta - \frac{\pi}{3})} \equiv 0$

Proof:

$\text{LHS} \equiv \cos{\theta} - (2\cos{\theta}\cos{\frac{\pi}{3}}) \text{(by Sums to Products)}$

$\text{LHS} \equiv \cos{\theta} - \cos{\theta} \equiv 0 \equiv \text{RHS}$

Now, let the integral in question be $I$ .

Then from Identity 1, $8I = \int_0^\pi \cos{(4x + 4\sin{3x})} - 4\cos{(2x + 2\sin{3x})} + 3 \, \text{d}x$

Break off the constant term and evaluate it directly to obtain

$8I = 3\pi + J$

Where $J = \int_0^\pi \cos{(4x + 4\sin{3x})} - 4\cos{(2x + 2\sin{3x})} \, \text{d}x$

Use the standard border flip or substitute $u = \pi-x$ and we have

$J = \int_0^\pi \cos{(4x - 4\sin{3x})} - 4\cos{(2x - 2\sin{3x})} \, \text{d}x$

Add these two together and by direct expansion or Sums to Products, we obtain

$J = \int_0^\pi \cos{4x}\cos{(\sin{3x})} - 4\cos{2x}\cos{(2\sin{3x})} \, \text{d}x$

Break up $J$ along the three intervals: $\{0,\frac{\pi}{3}\} , \{\frac{\pi}{3},\frac{2\pi}{3}\} , \{\frac{2\pi}{3},\pi\}$

Leave the first interval alone.

For the second interval, substitute $u = x-\frac{\pi}{3}$

$J_2 = \int_0^\frac{\pi}{3} \cos{(4u+\frac{4\pi}{3})} \cos{(4\sin{(3u + \pi)})} - 4\cos{(2u+\frac{2\pi}{3})} \cos{(2\sin{(3u+\pi)})} \, \text{d}x$

$J_2 = -\int_0^\frac{\pi}{3} \cos{(4x+\frac{\pi}{3})}\cos{(4\sin{3x})} - 4\cos{(2x-\frac{\pi}{3})}\cos{(2\sin{3x})} \, \text{d}x$

For the third interval, substitute $u = x-\frac{2\pi}{3}$

$J_3 = \int_0^\frac{\pi}{3} \cos{(4u+\frac{8\pi}{3})} \cos{(4\sin{(3u + 2\pi)})} - 4\cos{(2u+\frac{4\pi}{3})} \cos{(2\sin{(3u+2\pi)})} \, \text{d}x$

$J_3 = -\int_0^\frac{\pi}{3} \cos{(4x-\frac{\pi}{3})}\cos{(4\sin{3x})} - 4\cos{(2x+\frac{\pi}{3})}\cos{(2\sin{3x})} \, \text{d}x$

Adding up all three pieces of the integral together and using Identity 2, we find at last that $J = 0$

Hence, $8I = 3\pi$

And finally, we have $a+b = 3+8 = 11$