# Can someone help me solve this using substitution please?

• Feb 6th 2010, 07:20 AM
s3a
Can someone help me solve this using substitution please?
I tried to get the detailed solution from Wolfram Alpha: http://www.wolframalpha.com/input/?i=integral+((tanx)^3)

but, it uses the "reduction formula" which is something I have not learned yet not to mention that this question is in the "substitution" section of my text book.

I was just hoping somebody could solve this for me using substitution because when I let u = tanθ, I get stuck and I can't see what else to do.

Any help would be GREATLY appreciated!

P.S.
If you want to see my work, it is attached. This particular problem is the last one of the PDF and is #47.
• Feb 6th 2010, 08:00 AM
skeeter
Quote:

Originally Posted by s3a
I tried to get the detailed solution from Wolfram Alpha: http://www.wolframalpha.com/input/?i=integral+((tanx)^3)

but, it uses the "reduction formula" which is something I have not learned yet not to mention that this question is in the "substitution" section of my text book.

I was just hoping somebody could solve this for me using substitution because when I let u = tanθ, I get stuck and I can't see what else to do.

Any help would be GREATLY appreciated!

P.S.
If you want to see my work, it is attached. This particular problem is the last one of the PDF and is #47.

$\tan^3{t} = \tan{t} \cdot \tan^2{t} = \tan{t}(\sec^2{t} - 1) = \tan{t}\sec^2{t} - \tan{t}$

$\int \tan{t}\sec^2{t} - \tan{t} \, dt = \int \tan{t} \sec^2{t} \, dt - \int \tan{t} \, dt$

can you finish?
• Feb 6th 2010, 08:03 AM
deathbycalc
All you have to do is integrate for u and evaluate [F(b) - F(a)]

(U^4)/4

[F(√3) - F(-√3)]

...to solve it from your PDF work that is
• Feb 6th 2010, 08:14 AM
skeeter
Quote:

Originally Posted by deathbycalc
All you have to do is integrate for u and evaluate [F(b) - F(a)]

(U^4)/4

[F(√3) - F(-√3)]

...to solve it from your PDF work that is

the pdf work is incorrect.
• Feb 6th 2010, 09:31 AM
s3a
@deathbycalc: my last step was incorrect because I still had du/dθ = (secθ)^2.

@skeeter: I completed it and got the right answer but could you just check if it's not just a fluke please?
• Feb 6th 2010, 09:40 AM
skeeter
Quote:

Originally Posted by s3a
@deathbycalc: my last step was incorrect because I still had du/dθ = (secθ)^2.

@skeeter: I completed it and got the right answer but could you just check if it's not just a fluke please?

for future reference, note that ...

$\int_{-a}^a (an \, odd \, function) \, dx = 0$