Hi all,

I am stuck on trying to find $\displaystyle \int sin(2x+\frac{\pi}{4})cos(2x+\frac{\pi}{4})dx$

Printable View

- May 11th 2011, 02:59 PMOilerStuck on finding an Integral
Hi all,

I am stuck on trying to find $\displaystyle \int sin(2x+\frac{\pi}{4})cos(2x+\frac{\pi}{4})dx$ - May 11th 2011, 03:06 PMTKHunny
Have you considered $\displaystyle \sin(2x) = 2\cdot\sin(x)\cos(x)$?

Mustn't forget one's trigonometry. - May 11th 2011, 03:56 PMChaobunny
Try u substitution, with $\displaystyle u=sin(2x+\pi/4)$. Then $\displaystyle du=2cos(2x+\pi/4)$, and the rest is easy.

- May 11th 2011, 03:59 PMOiler
Not sure how I would rewrite the expression using that Identity - something like $\displaystyle \frac{1}{4}\int sin(2(x+\frac{\pi}{2})) dx$ ?

- May 11th 2011, 04:05 PMOiler
- May 11th 2011, 04:26 PMOiler
Would it then be ...$\displaystyle \frac{1}{2}\int u du = \frac{-1}{2}*\frac{u^2}{2} = \frac{-1}{4}cos(2x+\frac{\pi}{4})$ ?

- May 11th 2011, 05:28 PMTKHunny
Unique solutions don't care how you find them. Good work, excepting the factor of 2 you missed.

Wonderful exercise: If you do it with the trig substitution, you get the sine, but squared. Can you demonstrate that BOTH are correct?

Hint: There's something about the solution of an indefinite integral that you probably think very little about that makes a difference. - May 11th 2011, 06:16 PMmathprofessor
Before using the excellent hint above try using the substitution u =(2x + pi/4). It's not mandatory to make this substitution but I've always told my students to let the angle equal U, unless the angle is just x or just t, etc or if there are two angles in the integrand (like your) and they are different angles (unlike yours). This make the integral look much nicer and gives you a better chance to see that the hint above is one way to go.

Good luck

Steven