Question

0. $\int sin^4x dx$

Answer 1

$\frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$

Answer 2

The problem asks to evaluate the integral $\int \sin^4x \, dx$.

To integrate even powers of sine or cosine, we use power-reducing trigonometric identities. We know the identity: $\sin^2x = \frac{1 - \cos 2x}{2}$

Now, we can express $\sin^4x$ as $(\sin^2x)^2$: $\sin^4x = \left(\frac{1 - \cos 2x}{2}\right)^2$ $= \frac{1}{4}(1 - \cos 2x)^2$ Expand the squared term: $= \frac{1}{4}(1 - 2\cos 2x + \cos^2 2x)$ We have a $\cos^2 2x$ term, which is also an even power. We need to reduce its power using another identity: $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ Here, $\theta = 2x$, so: $\cos^2 2x = \frac{1 + \cos(2 \cdot 2x)}{2} = \frac{1 + \cos 4x}{2}$ Substitute this back into the expression for $\sin^4x$: $\sin^4x = \frac{1}{4}\left(1 - 2\cos 2x + \frac{1 + \cos 4x}{2}\right)$ To combine the terms inside the parenthesis, find a common denominator: $= \frac{1}{4}\left(\frac{2}{2} - \frac{4\cos 2x}{2} + \frac{1 + \cos 4x}{2}\right)$ $= \frac{1}{4}\left(\frac{2 - 4\cos 2x + 1 + \cos 4x}{2}\right)$ $= \frac{1}{8}(3 - 4\cos 2x + \cos 4x)$ Now, we can integrate this simplified expression term by term: $\int \sin^4x \, dx = \int \frac{1}{8}(3 - 4\cos 2x + \cos 4x) \, dx$ $= \frac{1}{8} \int (3 - 4\cos 2x + \cos 4x) \, dx$ Integrate each term:

1. $\int 3 \, dx = 3x$
2. $\int -4\cos 2x \, dx = -4 \left(\frac{\sin 2x}{2}\right) = -2\sin 2x$
3. $\int \cos 4x \, dx = \frac{\sin 4x}{4}$

Combine these results and add the constant of integration, C: $\int \sin^4x \, dx = \frac{1}{8} \left( 3x - 2\sin 2x + \frac{\sin 4x}{4} \right) + C$ Distribute the $\frac{1}{8}$: $= \frac{3x}{8} - \frac{2\sin 2x}{8} + \frac{\sin 4x}{32} + C$ $= \frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$

The final answer is $\boxed{\frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C}$.