# Find $\int_0^\pi \cos^4\theta \sin^3\theta~d\theta$ using de Moivre's theorem

Find $$\displaystyle\int_0^\pi \cos^4\theta \sin^3\theta~d\theta$$ using de Moivre's theorem.

So I need to find and expression for $$\cos^4\theta \sin^3\theta$$ in terms of multiple angles. I know that $$2\cos\theta = z + z^{-1}$$ and $$2i\sin\theta = z-z^{-1}$$ and my original thought was to work out $$\cos^4\theta$$ and $$\sin^3\theta$$ and then multiply my two expressions but I've now realised that this is something I won't be able to easily integrate. In addition to this the expression for $$\cos^4\theta \sin^3\theta$$ can be expression entirely as multiple angles of $$\sin$$. This has left me unsure on working out a expression for $$\cos^4\theta \sin^3\theta$$.

Any help would be appreciated

• Your original thought was correct. You'll get a combination of powers of $z$, which are easy to integrate. – Robert Israel Nov 4 '18 at 20:26
• Sorry but not using the De Moivre's theorem. $$\int_0^{\pi} \sin^3 x\cos^4 x dx=2\int_0^{\pi/2} \sin^3 x\cos^4 x dx=B\left(\frac 52,2\right)=\frac {4}{35}$$ – Rohan Shinde Nov 6 '18 at 3:18

Indeed, we have $$\cos^4\theta\sin^3\theta=\frac{3\sin\theta + 3 \sin 3\theta - \sin 5\theta - \sin 7\theta}{64}$$ which you can deduce from de Moivre's $$(\cos x+i\sin x)^n=\cos nx+i\sin nx$$ and Euler's $$e^{ix}=\cos\theta+i\sin x$$ or just by repeated product-to-sum. Now integrate from 0 to $$\pi$$.
• but how did you get the expression for $\cos^4\theta \sin^3\theta$ as that is what I'm stuck on – H.Linkhorn Nov 4 '18 at 20:35
• Multiply $16\cos^4\theta=(e^{i\theta}+e^{-i\theta})^2=e^{i4\theta}+4e^{i2\theta}+6+4e^{-i2\theta}+e^{-i4\theta}$ and $-8i\sin^3\theta=(e^{i\theta}-e^{-i\theta})^3=e^{i3\theta}-3e^{i\theta}+3e^{-i\theta}-e^{-i3\theta}$. – user10354138 Nov 4 '18 at 20:42
• No. You get $-2^7i\cos^4\theta\sin^4\theta=(e^{i7\theta} - e^{-i7\theta}) + (e^{i5\theta} - e^{-i5\theta}) - 3 (e^{i3\theta} -e^{-i3\theta}) - 3 (e^{i\theta} - e^{-i\theta})$ – user10354138 Nov 4 '18 at 20:55
Use the properties that $$\cos^2\theta =\frac{1+\cos2\theta}{2}$$ $$\sin^2\theta=\frac{1-\cos2\theta}{2}$$ $$\cos\theta \sin\alpha= \frac{1}{2} [\sin(\theta+\alpha)+\sin(\theta-\alpha)]$$ $$\cos\theta \cos\alpha= \frac{1}{2} [\cos(\theta+\alpha) +\cos(\theta-\alpha)]$$ $$\sin\theta \sin\alpha= \frac{1}{2} [\cos(\theta-\alpha)-\cos(\theta+\alpha)]$$