# Find the function f whose derivative is $\sin^2 x$ [closed]

Find the function $f$ whose derivative is

$$f'(x)=\sin^2(x)$$

and where $f(\pi)=\pi$.

$$f(x)=\frac12(x-\sin x\cos x)$$ What do I do from here?

• Make sure that letting $x=\pi$ makes $f(x)=\pi$. Note that the antiderivative of $f'(x)$ is $f(x)+C$ for some constant $C$. – abiessu Sep 10 '13 at 15:12

You label this as differential equations, is this supposed to be integral calculus?

You have solved the indefinite integral and forgot to add +c.

$$\int \sin^2(x)dx=\int \frac{1-\cos(2x)}{2}dx=\frac{x}{2}-\frac{\sin(2x)}{4}+c.$$

Now evaluate at $f(\pi)=\pi$ to see $\pi=\frac{\pi}{2}+c$ and solve.

$$f'(x)=\sin^2x\implies f(x)=\frac12(x-\sin x\cos x) \color{red}{+ C}\;,\;\;C=\;\text{a constant} .$$
$$\pi=f(\pi)=\frac{\pi}2+C\implies C=\frac{\pi}2\implies \color{blue}{f(x)=\frac12(x-\sin x\cos x)+\frac{\pi}2}$$
$$f'(x) = \sin^2(x) \Rightarrow f(x) = \int \sin^2 x \ dx = \int \frac{1}{2} \left(1 - \cos 2x \right) \ dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$$
$$f(\pi) = \pi \Rightarrow \frac{\pi}{2}+ C = \pi\Rightarrow C = \frac{\pi}{2}$$
$$\Rightarrow f(x) = \frac{x}{2} - \frac{\sin 2x}{4} + \frac{\pi }{2}$$