# Math Help - [SOLVED] Definite Integral Problem

1. ## [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?

2. Originally Posted by fattydq
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?

3. Originally Posted by Chris L T521
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?

4. Originally Posted by fattydq
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?