Could someone please verify if the following two functions are equivalent (click to see).

Question

babyslapmafro · Answer

$$\frac{ \tan^4(4x)+2\tan^2(4x) }{ 16 }=\frac{ 1 }{ 16 }\sec^4(4x)$$

babyslapmafro · Answer

I was integrating and my answer is on the left, the answer in the book is on the right.

babyslapmafro · Answer

This is the integral
$$\int\limits_{}^{}\tan(4x)\sec^4(4x)dx$$

zehanz · Answer

Doesn't look the same.
First leave out the unimportant bits: set 4x =x and do not divide by 16.

Then it would have to be that: $\tan^4x+2\tan^2x=\sec^4x$

babyslapmafro · Answer

$$u=\tan(4x)$$
$$\int\limits_{}^{}(u)\sec^2(4x)(u^2+1)\frac{ du }{ 4\sec^2(4x) }$$
$$=\frac{ 1 }{ 4 }\int\limits_{}^{}(u^3+u)du$$

babyslapmafro · Answer

$$=\frac{ 1 }{ 4 }[\frac{ u^4 }{ 4 }+\frac{ u^2 }{ 2 }]+C$$

babyslapmafro · Answer

$$=\frac{ \tan^4(4x) }{ 16 }+\frac{ \tan^2(4x) }{ 8 }+C$$

babyslapmafro · Answer

=(my original answer as-in my first post)

babyslapmafro · Answer

yay or nay

zehanz · Answer

u=tan(4x), so du=4sec(4x).
You put in a factor du/(4sec(4x)), so that is 1. That is OK.
You also have a factor u²+1 = tan²(4x)+1=sec²(4x). I go with that one too.

Everything seems to be OK there!

babyslapmafro · Answer

ok so my answer is good?
i tried punching both functions in my calc and got different answers, i was careful with the input.

zehanz · Answer

I would rather say: I do not see where you made a mistake...
There is, somewhere.
Below I made a graph of $\tan^4(x)+2\tan^2(x)$ (blue) together with $\sec^4(x)-1$ (red, dotted).
So it seems your solution differs only a constant (-1) from the one in the book.
Well, that is the end of the problem: you just need another integration constant!

zehanz · Answer

In other words, altough both solutions (without the integration constant!) are different, they only differ 1, so you have made no mistake after all. Congratulations!

babyslapmafro · Answer

ok thanks

zehanz · Answer

YW!