Integrating $\sin x\cos x$

Question

Using substitution with $u=\sin{x}$, The integral of $\sin{x}\cos{x}$ is equal to $\frac{1}{2}\sin^2{x}$. This is the solution I require.

However, why is it that when I use the double angle formula $\sin{2x} = 2\sin{x} \cos{x}$, and integrate $\frac{1}{2}\sin{2x}$, I get a completely different answer of $-\frac{1}{4}\cos{2x}$ even though $\sin{2x} = 2\sin{x} \cos{x}$.

Note that $\cos(2x) = 1-2\sin^{2}(x)$. Substituting this in you can see that the the answers agree when you remember to account for the missing constant of integration. — JessicaK, May 20 '18 at 10:59
They are all the same but differ by a constant. Try graphing them to see it — The Integrator, May 20 '18 at 11:00

score 3 · Answer 1 · answered May 20 '18 at 11:00

3

It's actually not a completely different answer. By the double-angle formulae for cosine and the pythagorean identity, you have $$ \cos{2x} = \cos^2{x}-\sin^2{x} = 1-2\sin^2{x}, $$ so $-\frac{1}{4}\cos{2x}$ differs from $\frac{1}{2}\sin^2{x}$ by a constant. Since an indefinite integral is only determined up to a constant anyway, this is fine.

answered May 20 '18 at 11:00

Chappers

67,606

In fact you can do that integral in 6 ways and every answer would be the same – MR ASSASSINS117 May 20 '18 at 11:49
1

There're probably infinitely many ways, but yes, the Fundamental Theorem of Calculus implies they all differ by a constant. – Chappers May 20 '18 at 11:56

Integrating $\sin x\cos x$

1 Answers1