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Trig Integrals - How is what I am doing wrong?

  • Thread starter MathHawk
  • Start date
  • #1
9
0
I like to work these problems out and then check them with online integral calculators.

Homework Statement


[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan2x * sec4x dx



Homework Equations


[tex]\frac{d}{dx}[/tex]tanx = sec2x
sec2x = 1 + tan2x



The Attempt at a Solution



This seems so simple, using the identities and u substitution:

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan2x * sec4x dx
[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan2x * sec2x * sec2x dx
[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan2x * (tan2x + 1) * sec2x dx
[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex](tan4x + tan2x) * sec2x dx

Now: u = tanx, du = sec2x dx. tan [tex]\pi/4[/tex] = 1, tan 0 = 0/

[tex]\int[/tex][tex]^{1}_{0}[/tex](u4 + u2) du

= [[tex]\frac{1}{5}[/tex]u5 + [tex]\frac{1}{3}[/tex]u3][tex]^{1}_{0}[/tex]

[tex]\frac{1}{5}[/tex] + [tex]\frac{1}{3}[/tex] - (0 + 0) = [tex]\frac{8}{15}[/tex]


Therefore:

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan2x * sec4x dx = [tex]\frac{8}{15}[/tex]



Online indefinite integral calculators disagree with my indefinite integral, and the definite integral calculator I tried timed out. This looks flawless to me, but apparently I'm an idiot.

Thank you very much in advance for any help.
 
Last edited:

Answers and Replies

  • #2
33,154
4,838
I like to work these problems out and then check them with online integral calculators.

Homework Statement


[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan2x * sec4x dx



Homework Equations


[tex]\frac{d}{dx}[/tex]tanx = sec2x
sec2x = 1 + tan2x



The Attempt at a Solution



This seems so simple, using the identities and u substitution:

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan2x * sec4x dx
[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan2x * sec2x * sec2x dx
[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan2x * (tan2x + 1) * sec2x dx
The expression below does not follow from the one above.
[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan4x * tan2x * sec2x dx

Now: u = tanx, du = sec2x dx. tan [tex]\pi/4[/tex] = 1, tan 0 = 0/

[tex]\int[/tex][tex]^{1}_{0}[/tex](u4 + u2) du

= [[tex]\frac{1}{5}[/tex]u5 + [tex]\frac{1}{3}[/tex]u3][tex]^{1}_{0}[/tex]

[tex]\frac{1}{5}[/tex] + [tex]\frac{1}{3}[/tex] - (0 + 0) = [tex]\frac{8}{15}[/tex]


Therefore:

[tex]\int[/tex][tex]^{\pi/4}_{0}[/tex]tan2x * sec4x dx = [tex]\frac{8}{15}[/tex]



Online indefinite integral calculators disagree with my indefinite integral, and the definite integral calculator I tried timed out. This looks flawless to me, but apparently I'm an idiot.

Thank you very much in advance for any help.
 
  • #3
9
0
You're right. However, I don't have that on paper (I have the + in place of the *). It's corrected on here when I use u substitution, and it doesn't affect my answer. Is there anything else, or did I just whoop a computer integral calculator (doubtful)?
 
  • #4
You apparently don't know how to use a computer integral calculator because your answer (and procedure) is correct.
 

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