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Math Help - [SOLVED] Integral of Trig functions.

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    Ife
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    [SOLVED] Integral of Trig functions.

    How do i integrate  2{\sqrt{sin4x}}?
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    Ife
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    Quote Originally Posted by General View Post
    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 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 \sqrt{sin 4x}+1 and an outer radius of 2. So the volume would be: (\int{3- 2\sqrt{sin4x} - sin 4x} ) dx is this correct? From there i was stuck on the integral of the middle term...
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    Quote Originally Posted by Ife View Post
    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 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 \sqrt{sin 4x}+1 and an outer radius of 2. So the volume would be: (\int{3- 2\sqrt{sin4x} - sin 4x} ) dx is this correct? From there i was stuck on the integral of the middle term...
    V = \pi \int_0^{\frac{\pi}{8}} 2^2 - (\sqrt{\sin(4x)} + 1)^2 \, dx

    get out your calculator to find V ... I don't believe it was intended that you complete this evaluation by hand.
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    Ife
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    Quote Originally Posted by skeeter View Post
    V = \pi \int_0^{\frac{\pi}{8}} 2^2 - (\sqrt{\sin(4x)} + 1)^2 \, dx

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