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Math Help - [SOLVED] Definite Integral Problem

  1. #1
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    [SOLVED] Definite Integral Problem

    I'm asked to evaluate the definite integral of the square root of 1+cos(4x)dx from 0 to pi/4. I don't even know which method would be best to use here, so essentially I don't even know how to start this problem! Could somebody help me out?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by fattydq View Post
    I'm asked to evaluate the definite integral of the square root of 1+cos(4x)dx from 0 to pi/4. I don't even know which method would be best to use here, so essentially I don't even know how to start this problem! Could somebody help me out?
    Hint: \sqrt{1+\cos(4x)}=\frac{\sqrt{2}}{\sqrt{2}}\sqrt{1  +\cos(4x)}=\sqrt{2}\sqrt{\frac{1+\cos(4x)}{2}}=\sq  rt{2}\cos\!\left(\frac{4x}{2}\right)=\sqrt{2}\cos(  2x) by applying the half angle identity \cos\!\left(\frac{\alpha}{2}\right)=\sqrt{\frac{1+  \cos\alpha}{2}}

    So \int_0^{\frac{\pi}{4}}\sqrt{1+\cos(4x)}\,dx=\sqrt{  2}\int_0^{\frac{\pi}{4}}\cos(2x)\,dx, which is easier to integrate.

    Can you continue from here?
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    Quote Originally Posted by Chris L T521 View Post
    Hint: \sqrt{1+\cos(4x)}=\frac{\sqrt{2}}{\sqrt{2}}\sqrt{1  +\cos(4x)}=\sqrt{2}\sqrt{\frac{1+\cos(4x)}{2}}=\sq  rt{2}\cos\!\left(\frac{4x}{2}\right)=\sqrt{2}\cos(  2x) by applying the half angle identity \cos\!\left(\frac{\alpha}{2}\right)=\sqrt{\frac{1+  \cos\alpha}{2}}

    So \int_0^{\frac{\pi}{4}}\sqrt{1+\cos(4x)}\,dx=\sqrt{  2}\int_0^{\frac{\pi}{4}}\cos(2x)\,dx, which is easier to integrate.

    Can you continue from here?
    Ahh that's useful thanks, but how did you know to multiply it by sqrt2/sqrt2?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by fattydq View Post
    Ahh that's useful thanks, but how did you know to multiply it by sqrt2/sqrt2?
    I saw in the original problem that it had \sqrt{1+\cos(4x)}. This looked similar to \sqrt{1+\cos\alpha} if we take \alpha=4x. This now looks similar to the half angle equation for cosine, which is \cos\!\left(\frac{\alpha}{2}\right)=\sqrt{\frac{1+  \cos\alpha}{2}}. All I need to do with \sqrt{1+\cos\alpha} is to get a two in the denominator. Multiplying by \frac{\sqrt{2}}{\sqrt{2}} allows me to do this, since \frac{\sqrt{2}}{\sqrt{2}}\sqrt{1+\cos\alpha}=\sqrt  {2}\sqrt{\frac{1+\cos\alpha}{2}}. Now I can apply the identity to get \sqrt{2}\cos\!\left(\frac{\alpha}{2}\right) where we said \alpha=4x.

    Does this clarify things?
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