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Question
Math
Posted 3 months ago
I2=0π2sin4xdxI_{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 sin4(x) \sin^4(x) from 0 0 to π2 \frac{\pi}{2} , we use the power-reduction formula for even powers of sine
step 2
The integral of sin4(x) \sin^4(x) can be expressed as 0π23814cos(2x)+132cos(4x)dx \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 [3x818sin(2x)+1128sin(4x)]0π2 \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 0 and π2 \frac{\pi}{2} , we find (3π160+0)(00+0) \left(\frac{3\pi}{16} - 0 + 0\right) - \left(0 - 0 + 0\right)
step 5
The result of the definite integral is 3π16 \frac{3\pi}{16}
Answer
3π16 \frac{3\pi}{16}
Key Concept
Power-Reduction Formula for Even Powers of Sine
Explanation
The integral of sin4(x) \sin^4(x) from 0 0 to π2 \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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