# Is my book wrong?

I have this integration problem that I did but it doesnt give me the right answer. But there are like 3 other similar exercices I did the same way and I got all the right answers.. maybe it is my book (I dont think so ... :p)

It's an integration problem:

$$\int tg^3(4x) sec^4(4x)dx$$

heres what I did:

$$\int (sec^2(4x)-1)(tg(4x))(sec^4(4x))dx$$
= $$\int (sec^6(4x) - sec^4(4x)) (tg4x) dx$$

u= sec 4x
du = 4(sec4x)(tg4x)dx --> dx = du/(4(sec4x)(tg4x))

=$$\int \frac{(u^6 - u^4) du}{4u}$$

(replaced the sec(4x) at denom. with u since u=sec4x)

=$$\frac{1}{4}\int u^5 -$$ $$\frac{1}{4}\int u^3$$
= $$\frac{1}{24}sec^6(4x) -$$ $$\frac{1}{16} sec^4(4x)$$ +C

Add anything you want! Thanks everyone

BTW the answer in my book is
(1/16) tg^4 (4x) + (1/24) tg^6(4x)
I tried it in my calc with an x and it doesnt give the same answer.

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Imo
Actually, those aer the same answers (up to that constant, if you put it directly into your calculator, they differ already by a constant that doesn't matter in this type of integration...add 1/48 to your answer to get the other). They just went another direction in the integration.

What the hell is tg(x)?

dextercioby
Homework Helper
Here's another way.Denote the integral by I and write everything in terms of sine a cosine...

$$I=\int \left(\frac{\sin^{3}4x}{\cos^{7}4x}\right) \ dx$$

$$4x=u \Rightarrow 4dx=du$$

$$I=\frac{1}{4}\int \left(\frac{\sin^{4}u}{\cos^{7}u}\right) \ du =-\frac{1}{4}\int \left(\frac{1-\cos^{2}u}{\cos^{7}u}\right) \ d(\cos u) =-\frac{1}{4}\left[\frac{(\cos u)^{-6}}{-6}-\frac{(\cos u)^{-4}}{-4}\right] +\mathcal{C}$$

$$I=\frac{1}{24}\frac{1}{\cos^{6}4x}-\frac{1}{16}\frac{1}{\cos^{4}4x}+\mathcal{C}$$

Daniel.

dextercioby
Homework Helper
whozum said:
What the hell is tg(x)?
Tangent of (x),what else??? Daniel.

dextercioby said:
Here's another way.Denote the integral by I and write everything in terms of sine a cosine...

$$I=\int \left(\frac{\sin^{3}4x}{\cos^{7}4x}\right) \ dx$$

$$4x=u \Rightarrow 4dx=du$$

$$I=\frac{1}{4}\int \left(\frac{\sin^{4}u}{\cos^{7}u}\right) \ du =-\frac{1}{4}\int \left(\frac{1-\cos^{2}u}{\cos^{7}u}\right) \ d(\cos u) =-\frac{1}{4}\left[\frac{(\cos u)^{-6}}{-6}-\frac{(\cos u)^{-4}}{-4}\right] +\mathcal{C}$$

$$I=\frac{1}{24}\frac{1}{\cos^{6}4x}-\frac{1}{16}\frac{1}{\cos^{4}4x}+\mathcal{C}$$

Daniel.

I see my solution was okay. We had to do the problem by exponents of sec and tg. (sorry I dont know what the method is called in english)

However, what does d(cos u) means in your solution? and I know you replaced a sin^4 (u) by (1-cos^2 (u)), but where is the other sin^2 (x)
Surely it is the d(cos u) you used, but I'm not familiar with this notation

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dextercioby said:
Tangent of (x),what else??? Daniel.
What happened to tan(x)?

cepheid
Staff Emeritus