Find the antiderivative of f(x) = cos^2 x, and also its derivative.

Step-by-step solution

1. 1

   Replace the square with the power-reduction identity. Squares of sine and cosine cannot be integrated directly with the power rule, but they linearise:

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

   This comes from $\cos 2x=2\cos^2x-1$ rearranged, and it is the standard first move for $\int\cos^2$ or $\int\sin^2$.

2. 2

   Split the resulting integral.

   $\int\cos^2 x\,dx=\frac{1}{2}\int 1\,dx+\frac{1}{2}\int\cos 2x\,dx$

3. 3

   Integrate each piece. The second needs the $\tfrac1k$ factor for the inner $2x$:

   $\frac{1}{2}\int 1\,dx=\frac{x}{2},\qquad\frac{1}{2}\int\cos 2x\,dx=\frac{1}{2}\cdot\frac{\sin 2x}{2}=\frac{\sin 2x}{4}$

   $\int\cos^2 x\,dx=\frac{x}{2}+\frac{\sin 2x}{4}+C$

4. 4

   Differentiate $f$ as well, using the chain rule. Writing $f(x)=(\cos x)^2$:

   $f'(x)=2\cos x\cdot(-\sin x)=-2\sin x\cos x=-\sin 2x$

   where the last step uses the double-angle identity $\sin 2x=2\sin x\cos x$.

5. 5

   Check the antiderivative by differentiating it back.

   $\frac{d}{dx}\left(\frac{x}{2}+\frac{\sin 2x}{4}\right)=\frac{1}{2}+\frac{2\cos 2x}{4}=\frac{1+\cos 2x}{2}=\cos^2 x\qquad\checkmark$

   Numerically at $x=0.9$: the derivative of the antiderivative is $0.386399$ and $\cos^2(0.9)=0.386399$ $\checkmark$. The average value $\tfrac{x}{2}$ also matches the known mean of $\cos^2$ over a period, namely $\tfrac12$.