# Integrating Trig functions (Namely tan^{2}x)

1. Jan 14, 2007

### aFk-Al

I know that
$$\int \tan^{2}x dx= \int \sec^{2}x-1\ dx = \tan x - x + C$$
but i don't completely understand how this is derived. Because of this lack of comprehension, I have no idea what to do with $$\int \sec^{4}5x\ dx$$. I went from there to get:
$$\int \sec^{4}5x\ dx=$$
$$\int [\sec^{2}5x]^{2}\ dx=$$
$$\int \tan^{2}5x\ dx =$$
but I have no idea where to go from here. Any help would be appreciated. Thanks in advance.

Last edited: Jan 14, 2007
2. Jan 14, 2007

change from [text] to $$3. Jan 14, 2007 ### radou ...and change from [\text] to$$.

4. Jan 14, 2007

### aFk-Al

Ha there we go! Thank you. How about help with the problem now?

5. Jan 14, 2007

### d_leet

This is wrong, these two lines are not equivalent. What trigonometric identity do you know that relates secant and tangent?

6. Jan 14, 2007

### aFk-Al

Doesn't $$\int \sec^{2}x\ dx=\int \tan x dx$$?

Edit: I answered the question as I was asking it. It equals $$\tan x$$ not $$\int \tan x dx$$. So what should I do instead? I would also still like to know how to integrate $$\tan^2 x$$

Last edited: Jan 14, 2007
7. Jan 14, 2007

### d_leet

No. What is the derivative of tan(x)?

8. Jan 14, 2007

### aFk-Al

$$\sec^2 x$$. roger. so what instead?

9. Jan 14, 2007

### d_leet

Ok Starting with

$$\int \sec^{4} 5x dx$$

split this up to

$$\int (\sec^{2} 5x)(\sec^{2} 5x) dx$$

Then use an identity to simplify the first sec2(5x) and then it should be much simpler.

10. Jan 14, 2007

### robphy

Use complex numbers via
$$\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix}})$$
$$\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix}})$$
where $$i^2=-1$$
... and make use of
$$e^{ix}=\cos(x)+i\sin(x)$$ (to convert back)

[Integrating the exponential function is much easier.]

11. Jan 14, 2007

### d_leet

With sines and cosines this is easy, but with secants and tangents i think this would actually complicate things.

12. Jan 14, 2007

### mjsd

FYI, the only trick you have used here is the fact that
$$\tan^2 x = \sec^2 x -1$$
which is a trig identity derived from
$$\cos^2 x + \sin^2 x = 1$$
btw, knowing this is inconsequential to how you may solve
$$\int \sec^4 (5x)\;dx$$

as previous posters mentioned, split them up and use the identity once...your aim is to rewrite the expression in terms of things you can readily integrate like
$$\sec^2 x, \sec^2 x \tan^2 x$$

hope I haven't given away too much

13. Jan 15, 2007

### dextercioby

$$=\frac{1}{5}\int \frac{du}{\cos^{4}u}$$

where $u=5x$.

Now make the substitution

$$\tan u =y$$

Daniel.

14. Jan 15, 2007

### dickcruz

sec^4 (5x) dx
let 5x = y (which you can differentialte and substitue accordingly).

so lets integrate
I= sec^4 x dx.
= (sec^2x . sec^2x) dx
= ((1+tan^2x) .sec^2x)dx
= Sec^2 dx + sec^2x.tan^2x dx
= tan x + c + J

J = sec^2x.tan^2x dx

Now let tan x =t
therefore, sec^2xdx =dt
substituting
J = t^2 dt
J= t^3 / 3 + c

I=tan x + c + (t^3 / 3) + c
tan x +(tan^3x)/3 + constant of indefinate integration.

15. Jan 15, 2007

### ashrafmod

that is good

16. Jan 15, 2007

### aFk-Al

This is great guys, thank you all. I just need to memorize all those stupid trig identities and I think this will be a lot easier for me. Thanks again everyone!

-Al

17. Jan 15, 2007

### d_leet

Please don't post complete solutions to homework questions.

18. Jan 15, 2007

### dickcruz

You dont have to memorize the identitites.
I derive them when i need them

the basic ones are

Sin^2 x + cos^2 x= 1
Cos2x = Cos^2 - sin^2x
sin 2 x = 2. cosx. sinx

from the first one we can divide throught by cos and get values for
tan^2x +1=sec^x.
dividing by sin gives
1+cot^2x = csc^2x.

from the second one
you can substitute sin^2x or cos^2x as 1-cos^2x or 1-sin^2x
this is how how you integrate

19. Jan 15, 2007

### aFk-Al

thanks for the advice, i'll work on that.

20. Jan 15, 2007

### drpizza

to add to what dickcruz said, the cos2x formulas (there are 3) are very useful in integration:

take cos2x= 1 - 2sin^2x and solve for sin^2x
(or take cos2x = 2cos^2x - 1 and solve for cos^2x
This gives you the power reducing formulas which you'll use when you have even powers.
i.e. to integrate (cos^2 x)
you would first change it to the integral of (1 + cos2x)/2; a relatively simple integral to take care of.