# 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.