Very simple trigonometric integral

**fysikbengt** · June 24th 2011, 08:28 AM

Now you will all point at me and laugh. I tried to prove that $\int_{0}^{\pi}\sin^{3}\theta d\theta=\frac{4}{3}$ and all I get is
$\int_{0}^{\pi}\sin^{3}\theta d\theta=\left[3\sin^{2}\theta\cos\theta\right]_{0}^{\pi}=3*0^{2}*(-1)-3*0^{2}*1=0$

Where does it all go so wrong?

**Quacky** · June 24th 2011, 08:37 AM

Originally Posted by fysikbengt

Now you will all point at me and laugh. I tried to prove that $\int_{0}^{\pi}\sin^{3}\theta d\theta=\frac{4}{3}$ and all I get is
$\int_{0}^{\pi}\sin^{3}\theta d\theta=\left[3\sin^{2}\theta\cos\theta\right]_{0}^{\pi}=3*0^{2}*(-1)-3*0^{2}*1=0$

Where does it all go so wrong?

You've differentiated; you wanted to integrate. To integrate this, I would either write it as $Sin\theta(1-Cos^2(\theta))$ (and expand the brackets) or use some sort of double angle formula to manipulate.

**Prove It** · June 24th 2011, 08:49 AM

Originally Posted by fysikbengt

Now you will all point at me and laugh. I tried to prove that $\int_{0}^{\pi}\sin^{3}\theta d\theta=\frac{4}{3}$ and all I get is
$\int_{0}^{\pi}\sin^{3}\theta d\theta=\left[3\sin^{2}\theta\cos\theta\right]_{0}^{\pi}=3*0^{2}*(-1)-3*0^{2}*1=0$

Where does it all go so wrong?

Maybe the reason you're getting the wrong answer is because $\displaystyle \int{\sin^3{\theta}\,d\theta} \neq 3\sin^2{\theta}\cos{\theta}$ ...

**e^(i*pi)** · June 24th 2011, 08:52 AM

You can also use the identity $\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)$ which I suspect Quacky was alluding to since $\sin(\theta) \text{ and } \sin(a\theta)$ (where a is a constant) are simple to integrate.

You get bonus kudos for deriving that identity yourself

**Prove It** · June 24th 2011, 09:00 AM

Originally Posted by e^(i*pi)

You can also use the identity $\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)$ which I suspect Quacky was alluding to since $\sin(\theta) \text{ and } \sin(a\theta)$ (where a is a constant) are simple to integrate.

You get bonus kudos for deriving that identity yourself

I'll use the lazy derivation and leave the proper derivation for the OP

$\displaystyle \begin{align*} (\cos{\theta} + i\sin{\theta})^3 &= \cos^3{\theta} + 3i\cos^2{\theta}\sin{\theta} + 3i^2\cos{\theta}\sin^2{\theta} + i^3\sin^3{\theta} \\ &= \cos^3{\theta} - 3\cos{\theta}\sin^2{\theta} + i(3\cos^2{\theta}\sin{\theta} - \sin^3{\theta}) \\ &= \cos^3{\theta} - 3\cos{\theta}(1 - \cos^2{\theta}) + i[3(1 - \sin^2{\theta})\sin{\theta} - \sin^3{\theta}] \\ &= \cos^3{\theta} - 3\cos{\theta} + 3\cos^3{\theta} + i(3\sin{\theta} - 3\sin^3{\theta} - \sin^3{\theta}) \\ &= 4\cos^3{\theta} - 3\cos{\theta} + i(3\sin{\theta} - 4\sin^3{\theta}) \end{align*}$

But by DeMoivre's Theorem, we also have

$\displaystyle (\cos{\theta} + i\sin{\theta})^3 = \cos{3\theta} + i\sin{3\theta}$ .

Equating real and imaginary parts gives

$\displaystyle \cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}$

and

$\displaystyle \sin{3\theta} = 3\sin{\theta} - 4\sin^3{\theta}$

**fysikbengt** · June 24th 2011, 10:01 AM

Originally Posted by e^(i*pi)

You can also use the identity $\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)$ which I suspect Quacky was alluding to since $\sin(\theta) \text{ and } \sin(a\theta)$ (where a is a constant) are simple to integrate.

You get bonus kudos for deriving that identity yourself

Sure! By looking at the unit circle you can prove that $\cos (x+y)=\cos x \cos y -\sin x \sin y$ and by substituting $x$ with $x-\pi /2$ this leads to $\sin(x+y)=\sin x \cos y +\cos x \sin y$ . From these two formulae you can easily prove that $\sin(2x)=2\sin x\cos x$ and $\cos(2x)=1-2\sin^{2}x$ .

Now the rest follows:
$\sin(3x)=\sin(x+2x)=\sin x\cos(2x)+\cos x\sin(2x)=\sin x(1-2\sin^{2}x)+\cos x 2\sin x\cos x=\sin x -2\sin^{3}x+2\sin x (1-\sin^{2}x)=3\sin x -4\sin^{3}x$
And that's it.

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