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$\begingroup$

Find the exact value of the following definite integral:
$\int ^\frac{\pi}{2}_{0} \sin\left(2x+\frac{\pi}{4}\right)\:dx=\left[-\frac{1}{2}(2x+\frac{\pi}{4})\right]^\frac{\pi}{2}_{0}$ $=-\frac{1}{2}\left(2\frac{\pi}{2}+\frac{\pi}{4}\right)+\frac{1}{2}\left(2\cdot 0+\frac{\pi}{4}\right)$ $=-\frac{1}{2}\left(\pi+\frac{\pi}{4}\right)+\frac{1}{2}\left(\frac{\pi}{4}\right)=-\frac{\pi}{2}$

but the right answer is:

$\int_{0}^{\frac{\pi}{2}}{\sin{\left(2x+\frac{\pi}{4}\right)\:dx}}=\frac{\sqrt{2}}{2}$

Help me out! thanks!

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    oh!....so sorry! my fault!!!!i didn't see.2012-07-27

2 Answers 2

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You have forget $\cos$ after you have done antiderivative.
Solve-

$\int^\frac{\pi}{2}_{0}\sin\left(2x+\frac{\pi}{4}\right)\:dx=\left[-\frac{1}{2}\cos\left(2x+\frac{\pi}{4}\right)\right]^\frac{\pi}{2}_{0}$ $=-\frac{1}{2}\cos\left(2\frac{\pi}{2}+\frac{\pi}{4}\right)+\frac{1}{2}\cos\left(2\cdot 0+\frac{\pi}{4}\right)$ $=-\frac{1}{2}\cos\left(\pi+\frac{\pi}{4}\right)+\frac{1}{2}\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \quad \blacksquare$

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$\int_0^{\pi/2}\sin(2x+\tfrac{\pi}{4})\,dx=[−\tfrac{1}{2}\cos(2x+\tfrac{π}{4})]_0^{\pi/2}=2\sqrt{2}$

  • 0
    [Please see here](http://meta.stackexchange.com/a/70559/161783) for how to typeset common math expressions with LaTeX, and [see here](http://meta.stackoverflow.com/editing-help) for how to use Markdown formatting.2012-07-29