# Thread: [SOLVED] Integral of Trig functions.

1. ## [SOLVED] Integral of Trig functions.

How do i integrate $\displaystyle 2{\sqrt{sin4x}}$?

2. Originally Posted by General
thanks... but i don't even understand that! i wonder if my calculations are correct? Let me give you the entire question:
Find the volume of the solid generated by revolving the region in the 1st quadrant bounded above by the line y=1, below by the curve $\displaystyle y= \sqrt{sin 4x}$, and on the left by the y-axis, about the line y=-1...

So i calculated this to be done by the washer method, with an inner radius of $\displaystyle \sqrt{sin 4x}+1$ and an outer radius of 2. So the volume would be: $\displaystyle (\int{3- 2\sqrt{sin4x} - sin 4x} ) dx$ is this correct? From there i was stuck on the integral of the middle term...

3. Originally Posted by Ife
thanks... but i don't even understand that! i wonder if my calculations are correct? Let me give you the entire question:
Find the volume of the solid generated by revolving the region in the 1st quadrant bounded above by the line y=1, below by the curve $\displaystyle y= \sqrt{sin 4x}$, and on the left by the y-axis, about the line y=-1...

So i calculated this to be done by the washer method, with an inner radius of $\displaystyle \sqrt{sin 4x}+1$ and an outer radius of 2. So the volume would be: $\displaystyle (\int{3- 2\sqrt{sin4x} - sin 4x} ) dx$ is this correct? From there i was stuck on the integral of the middle term...
$\displaystyle V = \pi \int_0^{\frac{\pi}{8}} 2^2 - (\sqrt{\sin(4x)} + 1)^2 \, dx$

get out your calculator to find $\displaystyle V$ ... I don't believe it was intended that you complete this evaluation by hand.

4. Originally Posted by skeeter
$\displaystyle V = \pi \int_0^{\frac{\pi}{8}} 2^2 - (\sqrt{\sin(4x)} + 1)^2 \, dx$

get out your calculator to find $\displaystyle V$ ... I don't believe it was intended that you complete this evaluation by hand.
oh, so my initial calculation was correct, then? thanks..