Can this be solved?

1. Feb 1, 2005

digink

I was given this problem and the persom whom I recieved it from told me they couldn't see how it would he solved or even if it had an answer. I think I might have come to a conclusion even thought it might be wrong, what do you guys think

$$\int tan^2x sec^3x dx$$

this is what i've been able to conclude, might be wrong but I gave it a shot

$$\int tan^2x sec^3x = \int (sin^2x/cos^2x)(1/cos^3x) dx$$
$$\int (sin^2x)(cos x)/(cos^3x) dx = \int sin^2x/cos^2x = \int tan^2x$$
$$= sec^2x + c$$

Im guessing thats probably wrong, but what do you guys think?

2. Feb 1, 2005

Muzza

You made a mistake when you "cancelled" the cosines.

$$\frac{\sin^2{x}}{\cos^2{x}} \cdot \frac{1}{\cos^3{x}} = \frac{\sin^2{x}}{\cos^5{x}}$$.

3. Feb 1, 2005

digink

oops for some reason my brain was thinking common denominator, ill try to see if I can solve it with what you just gave.. I figured if their was a mistake it was at that step.

edit: BTW can anyone else tell me if Muzza is right and I was wrong at that step? im thinking its correct but like I said some of you guys know a lot more then me.

Last edited: Feb 1, 2005
4. Feb 1, 2005

marlon

yes it can be solved

Muzza is correct...

marlon

Last edited: Feb 1, 2005
5. Feb 1, 2005

dextercioby

Since it is not in the HW section,i'll solve most of it...

$$I=\int \tan^{2}x \sec^{3}x dx=\int \frac{\sin^{2}x}{\cos^{5}x} dx$$ (1)

Use part integration to get:

$$I=\sin x \cdot (-1)\frac{\cos^{-4}x}{-4}-\frac{1}{4} \int \frac{dx}{\cos^{3}x} dx$$ (2)

Denote the last integral by J and it is computed as follows:

$$J=\int \frac{\cos x}{\cos^{4}x} dx=\int \frac{d(\sin x)}{(1-\sin^{2}x)^{2}}$$ (3)

Making the obvious substitution
$$\sin x =u$$ (4)
,"J" becomes:
$$J=\int \frac{du}{(1-u^{2})^{2}} =\int \frac{du}{(1+u)^{2}(1-u)^{2}}$$ (5)

Hopefully the decomposition into simple fractions that i made is correct:

$$\frac{1}{(1+u)^{2}(1-u)^{2}}=\frac{1}{4}\frac{u}{(1+u)^{2}}+\frac{1}{2}\frac{1}{(1+u)^{2}}-\frac{1}{4}\frac{u}{(1-u)^{2}}+\frac{1}{2}\frac{1}{(1-u)^{2}}$$ (6)

I'll let you compute the 4 elementary integrals that would yield "J" in terms of "u" and then invert the substitution (4) and finally compute "I" from (2).

Good luck!!

Daniel.

Last edited: Feb 1, 2005
6. Feb 1, 2005

arildno

An alternative to Daniel's approach, is to use the $$u=tan(\frac{x}{2})$$ substitution.
Since you chose not to post in the homework section, I won't give you the solution..

7. Feb 1, 2005

dextercioby

Guess what,Arildno...Wolfram Integrator's Mathematica 5.0 reccomends the same "tangent of half-angle" substitution... :tongue2:

Did u have a contribution at that software's birth...??

Daniel.

P.S.I knew about the solution Mathematica offered before "cooking" the solution... But i found it :yuck:

8. Feb 1, 2005

digink

Hey dexter im sorry but you sort of lost me at step 2 when you did the integration by parts. Were you using this formula(this is off the top of my head might be slightly different):

$$\int f(x)g'(x) dx = f(x)g(x) - \int f'(x)g(x) dx ?$$

Also im not 100% sure how to implement this at the moment but if the top if the direvative of the bottom cant you implement the natural log (1/u)? im not sure if that can be done with this case.

Thanks.

9. Feb 1, 2005

dextercioby

Yes,of course...

Well,at the moment i used,no...But in the progress of computing the 4 simple integrals,yes,natural logarithm pops up...

Daniel.

10. Feb 1, 2005

DoubleMike

Couldn't the integral be solved more easily with substitution?

$\int (\sec^2x - 1)\sec^3x dx$

perhaps
$\int \sec^5x - \sec^3x dx$

and then applying power reduction formulas... I'm not sure, because I don't know what they are off-hand, but I remember evaluating integrals similar to that by using subbing a simple trig identity as the first step.

Last edited: Feb 1, 2005
11. Feb 1, 2005

dextercioby

I don't know...Why don't u solve it and compare to my solution??And then with Mathematica's solution (half angle substitution)...

Daniel.

12. Feb 1, 2005

digink

that to me looks like it can also work, im feeling really lazy right now or i'd solve it, ill try it tomorrow.

Thanks for the assistance guys.