# Evaluating $\int ^\frac{\pi}{2}_{0} \sin\left(2x+\frac{\pi}{4}\right)\ dx$

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

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

Help me out! thanks!

-
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

You have forget $\cos$ after you have done antiderivative.
$$\int^\frac{\pi}{2}_{0}\sin\left(2x+\frac{\pi}{4}\right)\:dx=\left[-\frac{1}{2}\cos\left(2x+\frac{\pi}{4}\right)\right]^\frac{\pi}{2}_{0}$$ $$=-\frac{1}{2}\cos\left(2\frac{\pi}{2}+\frac{\pi}{4}\right)+\frac{1}{2}\cos\left(2\cdot 0+\frac{\pi}{4}\right)$$ $$=-\frac{1}{2}\cos\left(\pi+\frac{\pi}{4}\right)+\frac{1}{2}\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \quad \blacksquare$$
$$\int_0^{\pi/2}\sin(2x+\tfrac{\pi}{4})\,dx=[−\tfrac{1}{2}\cos(2x+\tfrac{π}{4})]_0^{\pi/2}=2\sqrt{2}$$