Why isn't the antiderivative of $\sin(2x) = \sin^2(x)$?

Question

I notice that derivative of $\sin^2(x) = 2\sin(x)\cos(x)$ which is also equal to $\sin(2x)$.

Now, according to the fundamental theorem of calculus, integration and differentiation are inverse processes, so the integration of $\sin(2x)$ should be $\sin^2(x) + C$ which I think is not true.

I thought that the reason for this may be because of the constant of integration. I noticed that, $$\int\sin(2x)\, dx = \frac{-\cos(2x)}{2} + C $$

Equating $\dfrac{-\cos(2x)}{2} + C = \sin^2(x)\implies \boxed{ C = \dfrac{\cos(2x)}{2} +\sin^2(x)}$.

Can anyone help me with this? I'm sorry if it's not clear.

This is why we tell students to know their trig identities before going to calculus. — Sean Roberson, Feb 21 '22 at 13:32

score 5 · Accepted Answer · answered Feb 21 '22 at 13:24

5

Use the formula ${\cos(2x)+1 \over 2} = \cos^2(x)$ to find the value of $C$.

Sidenote: the antiderivative of $\sin(2x)$ is indeed $ \sin^2(x)$.

answered Feb 21 '22 at 13:24

coudy

5,843

score 1 · Answer 2 · answered Feb 21 '22 at 14:12

1

Use identity $\cos 2x=\cos^2 x- \sin^2x=1-2\sin^2x$. Hence, $-\frac{\cos 2x}{2}=\sin^2x-\frac{1}{2}$

answered Feb 21 '22 at 14:12

Vasili

10,690

score 0 · Answer 3 · answered Feb 21 '22 at 14:15

0

You need to look at $$-\frac{\cos2x}{2} + C$$ Now, $\cos2x = 1 - 2\sin^{2}x$

Substituting gives

$$\dfrac{2\sin^{2}x-1}{2} + C$$

$$ = \sin^2x + C - \dfrac12$$

$$\sin^2x + D$$ where $ D = C - \dfrac12$

Hence the antiderivative of $\sin(2x) = \sin^2(x)$

answered Feb 21 '22 at 14:15

Muzammil ahmed

33
5

Why isn't the antiderivative of $\sin(2x) = \sin^2(x)$?

3 Answers3