# Math Help - Integration

1. ## Integration

Hello

Can someone help me with this please:

$\int\dfrac{3}{2}\cos^2(4x)$

I tried a number of things, but can't see what function differentiates to give $\cos^2(4x)$

A hint at the right approach would be very much appreciated.

Thank you.

2. ## Re: Integration

Hello, Furyan!

Can someone help me with this please?

. . $\int\tfrac{3}{2}\cos^2(4x)$

You need this identity: . $\cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}$

3. ## Re: Integration

Hi Furyan!

Are you aware of the double angle formula for $\cos(2y)$ in terms of $\cos^2y$?

4. ## Re: Integration

Originally Posted by Furyan
Hello

Can someone help me with this please:

$\int\dfrac{3}{2}\cos^2(4x)$

I tried a number of things, but can't see what function differentiates to give $\cos^2(4x)$

A hint at the right approach would be very much appreciated.

Thank you.
It would help if you realise that \displaystyle \begin{align*} \cos^2{X} \equiv \frac{1}{2} + \frac{1}{2}\cos{2X} \end{align*}...

5. ## Re: Integration

Originally Posted by Soroban
Hello, Furyan!

You need this identity: . $\cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}$

You don't really NEED this identity, though it is the simplest simplification.

Integration by Parts will also work...

6. ## Re: Integration

The following identity may be useful:

$\cos^2 x = \frac{1 + \cos 2x}{2}$

Try using it.

7. ## Re: Integration

Hello Soroban,

Originally Posted by Soroban
Hello, Furyan!

You need this identity: . $\cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}$

Thank you. I was able to find the integral using that identity and substituting $\cos^2 4\theta$ with $\dfrac{1}{2}(1 + \cos8\theta)$

8. ## Re: Integration

Hello Prove It

Originally Posted by Prove It
It would help if you realise that \displaystyle \begin{align*} \cos^2{X} \equiv \frac{1}{2} + \frac{1}{2}\cos{2X} \end{align*}...
Thank you. I am having trouble with that realisation but I'm working on it.

I will also try using integration by parts.

9. ## Re: Integration

Do you mean $\cos^2 4x = \frac{1 + \cos 8x}{2}$ instead of $\cos^2 4x = \frac{1+8 \cos x}{2}$?

10. ## Re: Integration

MarceloFantini

Originally Posted by MarceloFantini
Do you mean $\cos^2 4x = \frac{1 + \cos 8x}{2}$ instead of $\cos^2 4x = \frac{1+8 \cos x}{2}$?
Yes indeed, thank you for pointing that out. I have corrected my post accordingly. I did use the former when I found the integral.

11. ## Re: Integration

Originally Posted by Furyan
Hello Prove It

Thank you. I am having trouble with that realisation but I'm working on it.

I will also try using integration by parts.
Here's how that identity is derived.

First a proof of the angle sum formula \displaystyle \begin{align*} \cos{\left(\alpha + \beta\right)} \equiv \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)} \end{align*}

Ans 1

Now, using the angle sum formula with \displaystyle \begin{align*} \alpha = \beta = X \end{align*} we have

\displaystyle \begin{align*} \cos{\left(X + X\right)} &\equiv \cos{(X)}\cos{(X)} - \sin{(X)}\sin{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \sin^2{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \left[1 - \cos^2{(X)}\right] \textrm{ by the Pythagorean Identity} \\ \cos{(2X)} &\equiv 2\cos^2{(X)} - 1 \\ 2\cos^2{(X)} &\equiv 1 + \cos{(2X)} \\ \cos^2{(X)} &\equiv \frac{1}{2} + \frac{1}{2}\cos{(2X)} \end{align*}

And if you're trying to use integration by parts...

\displaystyle \begin{align*} I &= \int{\frac{3}{2}\cos^2{(4x)}\,dx} \\ I &= \frac{3}{2}\int{\cos{(4x)}\cos{(4x)}\,dx} \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} - \int{-4\sin{(4x)}\cdot \frac{1}{4}\sin{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{\sin^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1 - \cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1\,dx} - \int{\cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{8} \sin{(4x)} \cos{(4x)} + \frac{3}{2} \int{1\,dx} - \frac{3}{2} \int{ \cos^2{ (4x) }\,dx} \\ I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x - I \\ 2I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x \\ I &= \frac{3}{16}\sin{(4x)}\cos{(4x)} + \frac{3}{4}x + C \end{align*}

And with a little manipulation I'm sure you can get this to be the same form as your answer you gained using the double angle identity...

12. ## Re: Integration

Dear Prove It

Originally Posted by Prove It
Here's how that identity is derived.

First a proof of the angle sum formula \displaystyle \begin{align*} \cos{\left(\alpha + \beta\right)} \equiv \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)} \end{align*}

Ans 1

Now, using the angle sum formula with \displaystyle \begin{align*} \alpha = \beta = X \end{align*} we have

\displaystyle \begin{align*} \cos{\left(X + X\right)} &\equiv \cos{(X)}\cos{(X)} - \sin{(X)}\sin{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \sin^2{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \left[1 - \cos^2{(X)}\right] \textrm{ by the Pythagorean Identity} \\ \cos{(2X)} &\equiv 2\cos^2{(X)} - 1 \\ 2\cos^2{(X)} &\equiv 1 + \cos{(2X)} \\ \cos^2{(X)} &\equiv \frac{1}{2} + \frac{1}{2}\cos{(2X)} \end{align*}

And if you're trying to use integration by parts...

\displaystyle \begin{align*} I &= \int{\frac{3}{2}\cos^2{(4x)}\,dx} \\ I &= \frac{3}{2}\int{\cos{(4x)}\cos{(4x)}\,dx} \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} - \int{-4\sin{(4x)}\cdot \frac{1}{4}\sin{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{\sin^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1 - \cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1\,dx} - \int{\cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{8} \sin{(4x)} \cos{(4x)} + \frac{3}{2} \int{1\,dx} - \frac{3}{2} \int{ \cos^2{ (4x) }\,dx} \\ I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x - I \\ 2I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x \\ I &= \frac{3}{16}\sin{(4x)}\cos{(4x)} + \frac{3}{4}x + C \end{align*}

And with a little manipulation I'm sure you can get this to be the same form as your answer you gained using the double angle identity...
I have been looking at the double angle identity for $\cos2\theta$ and, for some reason, was having trouble realising what you had pointed out in your original post. Now it's crystal clear. Some how I keep missing the elementary principles.

I will try and find the integral again using the substitution you have suggested.

Thank you also for the help with using integration by parts. I'd never have been able to do that, but I am very interested in it and and will work through it until I understand it.

Thank you very much for the time you have taken to help me. It's extremely generous of you and very grateful.