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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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3  
hint: where is the $\cos$ after integration? –  Raymond Manzoni Jul 27 '12 at 11:41
    
oh!....so sorry! my fault!!!!i didn't see. –  Sb Sangpi Jul 27 '12 at 11:43

2 Answers 2

up vote 3 down vote accepted

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

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Please see here for how to typeset common math expressions with LaTeX, and see here for how to use Markdown formatting. –  Zev Chonoles Jul 29 '12 at 14:29

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