Fourth power of sine!

A Simpler Approach

We note that the integral is even, positive and symmetrical at $\frac \pi 2$ and $\frac {3 \pi}2$ , then we have:

$\begin{aligned} I & = \int_0^{2\pi} \sin^4 x \ dx \\ & = 4 \int_0^\frac \pi 2 \sin^4 x \ dx & \small \color{#3D99F6}{\text{Using the identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = 2 \int_0^\frac \pi 2 \sin^4 x + \cos^4 x \ dx \\ & = 2 \int_0^\frac \pi 2 (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x \ dx \\ & = 2 \int_0^\frac \pi 2 1 - \frac 12 \sin^2 2x \ dx \\ & = \pi - \frac 12 \int_0^\frac \pi 2 (1-\cos 4x) \ dx \\ & = \pi - \frac \pi 4 + \frac 18 \sin 4x \ \bigg|_0^\frac \pi 2 \\ & = \pi - \frac \pi 4 + 0 = \frac {3\pi}4 \end{aligned}$

$\implies a+b+c = 3+1+4 = \boxed{8}$

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Previous Solution

We note that the integral is even, positive and symmetrical at $\frac \pi 2$ and $\frac {3 \pi}2$ , then we have:

$\begin{aligned} I & = \int_0^{2\pi} \sin^4 x \ dx \\ & = 4 \int_0^\frac \pi 2 \sin^4 x \ dx \\ & = 2 \times 2 \int_0^\frac \pi 2 \cos^0 x \sin^4 x \ dx \\ & = 2 B \left(\frac 12, \frac 52 \right) & \small \color{#3D99F6}{B(\cdot) \text{ is beta function.}} \\ & = \frac {2 \Gamma \left(\frac 12 \right)\Gamma \left(\frac 52 \right)}{\Gamma \left(3 \right)} & \small \color{#3D99F6}{\Gamma(\cdot) \text{ is gamma function.}} \\ & = \frac {2 \sqrt \pi \cdot \frac 32 \cdot \frac 12 \sqrt \pi}{2!} \\ & = \frac {3 \pi}4 \end{aligned}$

$\implies a+b+c = 3+1+4 = \boxed{8}$

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References:

• Beta function
• Gamma function