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**Prove It** If you MUST use integration by parts, then rewrite your integral as

$\displaystyle \int{\sin^2{\theta}\,d\theta} = \int{1 - \cos^2{\theta}\,d\theta}$

$\displaystyle = \int{(1 - \cos{\theta})(1 + \cos{\theta})\,d\theta}$

Now apply integration by parts with $\displaystyle u = 1 - \cos{\theta}$ so $\displaystyle du = \sin{\theta}$ and $\displaystyle dv = 1 + \cos{\theta}$ so $\displaystyle v = \theta + \sin{\theta}$.

The integral becomes

$\displaystyle \int{(1 - \cos{\theta})(1 + \cos{\theta})\,d\theta} = (1 - \cos{\theta})(\theta + \sin{\theta}) - \int{\sin{\theta}(\theta + \sin{\theta})\,d\theta}$

$\displaystyle = \theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta} - \int{\sin^2{\theta}\,d\theta}$

So this gives

$\displaystyle \int{\sin^2{\theta}\,d\theta} = \theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta} - \int{\sin^2{\theta}\,d\theta}$

$\displaystyle 2\int{\sin^2{\theta}\,d\theta} = \theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta}$

$\displaystyle \int{\sin^2{\theta}\,d\theta} = \frac{1}{2}\left(\theta + \sin{\theta} - \theta\cos{\theta} - \sin{\theta}\cos{\theta} - \int{\theta\sin{\theta}\,d\theta}\right)$.

Apply integration by parts once more, then you will have your answer.