Write lorenz equation

Let’s compute the definite integral:

$$\int_{0}^{\pi} \sin^4 x \, dx$$

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Step 1: Use a power-reduction identity

We know that

$$\sin^2 x = \frac{1 – \cos(2x)}{2}$$

So,

$$\sin^4 x = (\sin^2 x)^2 = \left(\frac{1 – \cos(2x)}{2}\right)^2$$ $$\sin^4 x = \frac{1}{4}(1 – 2\cos(2x) + \cos^2(2x))$$

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Step 2: Simplify $\cos^2(2x)$

$$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$$

So,

$$\sin^4 x = \frac{1}{4}\left(1 – 2\cos(2x) + \frac{1 + \cos(4x)}{2}\right)$$ $$\sin^4 x = \frac{1}{4}\left(\frac{3}{2} – 2\cos(2x) + \frac{1}{2}\cos(4x)\right)$$ $$\sin^4 x = \frac{3}{8} – \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)$$

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Step 3: Integrate term by term

$$\int_{0}^{\pi} \sin^4 x \, dx = \int_{0}^{\pi} \left(\frac{3}{8} – \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)\right) dx$$ $$= \frac{3}{8}\int_{0}^{\pi} dx – \frac{1}{2}\int_{0}^{\pi}\cos(2x) dx + \frac{1}{8}\int_{0}^{\pi}\cos(4x) dx$$

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Step 4: Evaluate each integral

1. $\int_{0}^{\pi} dx = \pi$
2. $\int_{0}^{\pi} \cos(2x) dx = \frac{\sin(2x)}{2}\Big|_{0}^{\pi} = 0$
3. $\int_{0}^{\pi} \cos(4x) dx = \frac{\sin(4x)}{4}\Big|_{0}^{\pi} = 0$

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Step 5: Combine results

$$\int_{0}^{\pi} \sin^4 x \, dx = \frac{3}{8}\pi – 0 + 0 = \boxed{\frac{3\pi}{8}}$$

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✅ Final Answer:

$$\int_{0}^{\pi} \sin^4 x \, dx = \frac{3\pi}{8}$$

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