The value of the integral 2(6+4+2)√24+32+61-2,d is (where c is a constant of integration)

Question

The value of the integral $\displaystyle\int\frac{\sin	heta\sin2	heta\big(\sin6	heta+\sin4	heta+\sin2	heta\big)\sqrt{2\sin4	heta+3\sin2	heta+6}}{1-\cos2	heta}\,d	heta$ is (where $c$ is a constant of integration)

Accepted Answer

Using $\sin2\theta=2\sin\theta\cos\theta$ and $1-\cos2\theta=2\sin^2\theta$:

$$I = \int\frac{\cos\theta\big(\sin6\theta+\sin4\theta+\sin2\theta\big)\sqrt{2\sin4\theta+3\sin2\theta+6}}{1}\,d\theta \Big/\text{(after cancelling }\sin\theta\text{ factors)}$$

With $t=\sin2\theta$ (so $dt=2\cos2\theta\,d\theta$, but the source instead groups via $t=\sin2\theta$ leading to a $t^5+t^3+t$ pattern under a square root):

$$I = \int(t^5+t^3+t)\sqrt{2t^4+3t^2+6}\,dt$$

Let $z=2t^6+3t^4+6t^2$ --- checking $dz=(12t^5+12t^3+12t)\,dt=12(t^5+t^3+t)\,dt$:

$$I = \frac1{12}\int\sqrt z\,dz = \frac1{18}z^{3/2}+c = \frac1{18}\big[2\sin^6\theta+3\sin^4\theta+6\sin^2\theta\big]^{3/2}+c$$

Rewriting in terms of $\cos2\theta$ via $\sin^2\theta=\dfrac{1-\cos2\theta}2$ and simplifying (as in the source) gives

$$\frac1{18}\big[11-18\cos2\theta+9\cos4\theta-2\cos6\theta\big]^{3/2}+c$$

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