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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 $\cos\theta$?

3 Answers 3

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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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    @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 graph:http://www.wolframalpha.com/input/?i=sin%283x%29+from+-pi%2F2+to+pi%2F22012-12-27