Mastering Trig and Integrals for Success | Tips and Tricks Included!

[Sethka]

Trig and Integrals!

Hi there, this is my first post here and I'm hoping someone can help me out, I'm working on an assignment with integrals and while I can manage workng out the number versions of the question just fine I've now encountered Trig functions in integrals and I've hit a brick wall. Can anyone possibly help me get this stuff started? I need to know how to do this for my exam weeks next week. Thanks for any and all help.


tan^5x Sec^3X dx

Why must trigs be so painful?

 

[island-boy]

try to apply the trigonometic identity
tan^2x +1 =sec^2x

Also remember that the integral of secxtanx is secx and that the integral of sec^2x is tanx.

There are actually methods which are helpful in solving trigonometic identities (at least, I was taught them when I was taking the calculus series), I can't remember them all right now. It should be in your calc book.

hope that helps

 

[HallsofIvy]

Actually, my approach to almost any problem involving trig functions is to rewrite in terms of sine and cosine only!
$$tan^5(x) Sec^3(X)= \frac{sin^5(x)}{cos^5(x)}\frac{1}{cos^5(x)}= \frac{sin^5(x)}{cos^8(x)}$$
Since that involves an odd power of sin(x), Rewrite the integral as
[tex]\int\tan^5(x)sec^3(x)dx= \int\frac{sin^4(x)}{cos^8(x)}sin(x)dx=\int\frac{(1-cos^2(x))^4}{cos^8(x)}sin(x)dx[tex]
Now, let u= cos(x) so that du= -sin(x)dx and the integral becomes
$$\int\frac{(u^2- 1)^4}{u^8}du$$

 

[VietDao29]

Or as island-boy has pointed out. When you need to tackle an integral with the product of odd power of tan, and a sec (the power of sec can be even or odd) function you can use the fact that:
$$\frac{d}{dx} \sec x = \tan x \sec x$$
And the Trig Identity: tan²x + 1 = sec²x Or tan²x = sec²x - 1
It goes like this:
$$\tan ^ 5 x \sec ^3 x dx = \int (\tan ^ 4 x \sec ^ 2 x) (\tan x \sec x) dx$$
Now make the substitution: u = sec x, and see if you can finish the problem. :)

 

[HallsofIvy]

I was going to point out a stupid error, then realized I was the one who wrote it! Okay, so I'll just point out a typo!
I wrote:

$$\int\tan^5(x)sec^3(x)dx= \int\frac{sin^4(x)}{cos^8(x)}sin(x)dx=\int\frac{(1 -cos^2(x))^4}{cos^8(x)}sin(x)dx$$ Now, let u= cos(x) so that du= -sin(x)dx and the integral becomes$$\int\frac{(u^2- 1)^4}{u^8}du$$

In fact, since $1- cos^2(x)= sin^2(x)$, $sin^4(x)= (1- cos^2(x))^2$, not $(1- cos^2(x))^4$
Now, let u= cos(x) so that du= -sin(x)dx and the integral becomes$$\int\frac{(u^2- 1)^2}{u^8}du$$

 

[dextercioby]

I get

$$\int \tan^{5}x \sec^{3} \ dx =-\int \frac{\left(1-u^{2}\right)^{2}}{u^{8}} \ du$$ with $u=cos x$

which is a little different from what HofIvy wrote.

Daniel.