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When Evaluating the below definite integral $$\int_{0}^{\pi}(2\sin\theta + \cos3\theta)\,d\theta$$

I get this.$$\left [-2\cos\theta + \frac{\sin3\theta}{3} \right ]_{0}^{\pi} $$

In the above expression i see that $-2$ is a constant which was taken outside the integral sign while performing integration. Now the question is should $-2$ be distributed throughout or does it only apply to $\cos\theta$? This is what i mean. Is it $$-2\left[\cos(\pi) + \frac{\sin3(\pi)}{3} - \left ( \cos(0) + \frac{\sin3(0)}{3} \right ) \right]?$$ Or the $-2$ stays only with $\cos\theta$?

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up vote 3 down vote accepted

I don't know why you think $-2$ should be distributed throughout . The correct answer is $$\left [-2\cos \theta+\frac{\sin 3\theta}3\right]_0^{\pi}=-2\cos \pi+\frac{\sin 3\pi}{3}+2\cos 0-\frac{\sin 3\cdot 0}{3}$$ as you said. Indeed, $$\int_{0}^{\pi}(2\sin\theta + \cos3\theta)d\theta=2\int_{0}^{\pi}\sin\theta + \int_{0}^{\pi}\cos3\theta d\theta=-2\left [\cos \theta\right]_0^{\pi}+\left [\frac{\sin 3\theta}3\right]_0^{\pi}=\left [-2\cos \theta+\frac{\sin 3\theta}3\right]_0^{\pi}$$ It "only stays" with $\cos\theta$

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$$\int_{0}^{\pi}(2\sin\theta + \cos3\theta)d\theta=\left [-2\cos\theta + \frac{\sin3\theta}{3} \right ]_{0}^{\pi} $$ as you noted so $-2$ as you see in @Nameless's answer is just for cosine function. Not for all terms.

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Just think about the parity of the trigonometric functions and you're done $$\int_{0}^{\pi}(2\sin\theta + \cos3\theta)\,d\theta=2\int_{0}^{\pi}\sin\theta\,d\theta=4$$

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.What's the meaning of parity of trigonometric functions?(sorry i upvoted even before clarifying my doubt.) – alok Dec 27 '12 at 12:06
@alok: Note that $\int_{0}^{\pi}\cos3\theta\,d\theta=\int_{-\pi/2}^{\pi/2}\sin 3\theta\,d\theta$ and because $f(-\theta)=-f(\theta)$ the integral is $0$. Check its – user 1618033 Dec 27 '12 at 12:16

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