Can this integral be solved using substitution?

[digink]

I was given this problem and the persom whom I received 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 that's probably wrong, but what do you guys think?

 

[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}}.

 

[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? I am thinking its correct but like I said some of you guys know a lot more then me.

 

[marlon]

yes it can be solved

Muzza is correct...

marlon

 

[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.

 

[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..

 

[dextercioby]

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

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

 

[digink]

Hey dexter I am 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 I am not 100% sure how to implement this at the moment but if the top if the direvative of the bottom can't you implement the natural log (1/u)? I am not sure if that can be done with this case.

Thanks.

 

[dextercioby]

Hey dexter I am 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 ?

Yes,of course...

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

Thanks.

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

Daniel.

 

[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.

 

[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.

 

[digink]

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.

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

Thanks for the assistance guys.