Find the function $f$ whose derivative is


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

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

  • $\begingroup$ Make sure that letting $x=\pi$ makes $f(x)=\pi$. Note that the antiderivative of $f'(x)$ is $f(x)+C$ for some constant $C$. $\endgroup$ – 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.


Your answer is almost correct, yet it lacks the integration constant:

$$f'(x)=\sin^2x\implies f(x)=\frac12(x-\sin x\cos x) \color{red}{+ C}\;,\;\;C=\;\text{a constant} .$$

But we want

$$\pi=f(\pi)=\frac{\pi}2+C\implies C=\frac{\pi}2\implies \color{blue}{f(x)=\frac12(x-\sin x\cos x)+\frac{\pi}2}$$


$$ f'(x) = \sin^2(x) \Rightarrow f(x) = \int \sin^2 x \ dx = \int \frac{1}{2} \left(1 - \cos 2x \right) \ dx = \frac{x}{2} - \frac{\sin 2x}{4} + C $$

$$ f(\pi) = \pi \Rightarrow \frac{\pi}{2}+ C = \pi\Rightarrow C = \frac{\pi}{2} $$

$$ \Rightarrow f(x) = \frac{x}{2} - \frac{\sin 2x}{4} + \frac{\pi }{2} $$


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