Question

Math

Posted 3 months ago

`$I_{2}=\int_{0}^{\frac{\pi}{2}} \sin ^{4} x d x$`

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Answer from Sia

Posted 3 months ago

Solution by Steps

step 1

To find the integral of $\sin^4(x)$ from $0$ to $\frac{\pi}{2}$, we use the power-reduction formula for even powers of sine

step 2

The integral of $\sin^4(x)$ can be expressed as $\int_0^{\frac{\pi}{2}} \frac{3}{8} - \frac{1}{4} \cos(2x) + \frac{1}{32} \cos(4x) \, dx$

step 3

Integrating term by term, we get $\left[\frac{3x}{8} - \frac{1}{8} \sin(2x) + \frac{1}{128} \sin(4x)\right]_0^{\frac{\pi}{2}}$

step 4

Evaluating the antiderivative at the bounds $0$ and $\frac{\pi}{2}$, we find $\left(\frac{3\pi}{16} - 0 + 0\right) - \left(0 - 0 + 0\right)$

step 5

The result of the definite integral is $\frac{3\pi}{16}$

Answer

$\frac{3\pi}{16}$

Key Concept

Power-Reduction Formula for Even Powers of Sine

Explanation

The integral of $\sin^4(x)$ from $0$ to $\frac{\pi}{2}$ can be computed by using the power-reduction formula to rewrite the integral in terms of cosine, which simplifies the integration process.

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