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How to solve this simple integral?

$\int_a^b\cos\theta \sin\theta \,d\theta$

I tried integrating by parts: given the rule

$\int u dv = uv - \int v du$

I substituted $\cos \theta$ as $dv $ and $\sin \theta$ as $u$ but I end up obtaining

$0 = \sin \theta \sin \theta$

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    $\begingroup$ Integration by parts is nice. But we can use the substitution $u=\sin\theta$. Or else observe that $\sin\theta\cos\theta=\frac{1}{2}\sin 2\theta$. $\endgroup$ – André Nicolas Mar 5 '13 at 20:55
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Integration by parts works:

$u' = \cos \phi, v = \sin \phi, u = \sin \phi , v' = \cos \phi$ then

$$ \int_a^b u' v = \int_a^b \cos \phi \sin \phi d \phi = (\sin \phi)^2 - \int_a^b \sin \phi \cos \phi d \phi $$

Therefore

$$ 2 \int_a^b \cos \phi \sin \phi d \phi = \sin^2 \phi$$

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  • $\begingroup$ Thank you, I got a wrong sign $\endgroup$ – John Smith Mar 5 '13 at 21:03
  • $\begingroup$ @JohnSmith I thought you might. $\endgroup$ – tom b. Mar 5 '13 at 21:03
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You don't need integration by parts. Let $u=\sin\theta$ which means $du=\cos\theta d\theta$.

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    $\begingroup$ You don't need substitution either $\endgroup$ – L. F. Mar 5 '13 at 20:52
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Using integration by parts, you should actually get: $$\int \cos \theta\sin\theta\,d\theta = \sin^2\theta - \int\cos \theta\sin\theta\,d\theta$$ If $A=B-A$, then $A=B/2$. Here, $A$ is your integral, and $B$ is $\sin^2 \theta$.

There should be a constant in there, too, just for kicks.

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Using by parts, $\int_a^b\cos\theta\sin\theta d\theta=-\cos^2\theta|_a^b-\int_a^b\sin\theta\cos\theta d\theta\implies \int_a^b\cos\theta\sin\theta d\theta=\frac{1}{2}(\cos^2a-\cos^2b)$

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For any derivable function $\,f(x)\,$ , we have

$$\int f'(x)f(x)^ndx=\frac{f(x)^{n+1}}{n+1}+C\;\;,\;\;n\neq -1\Longrightarrow$$

$$\int\limits_1^b\sin\theta\cos\theta \,d\theta=\int\limits_a^b\sin\theta (\sin\theta)'\,d\theta=\left.\frac{\sin^2\theta}{2}\right|_a^b=\ldots$$

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