Calculus integration problem

[From: ] [author: ] [Date: 11-10-04] [Hit: ]

.x = (3/4)tanθSo...tanθ = (4x/3)θ = arctan(4x/3)So the integral is indeed........

x = (3/4)tan θ
dx = (3/4)(sec θ)^2 dθ

Now integrate...

∫[(3/4)(sec θ)^2]dθ/[9 + 16(9/16)(tan θ)^2]
(3/4)∫(sec θ)^2 dθ/[9(1 + (tanθ)^2)]
(3/36)∫(sec θ)^2 dθ/[(secθ)^2]
(1/12)∫dθ
(1/12)θ + C

Remember that...

x = (3/4)tanθ

So...

tanθ = (4x/3)
θ = arctan(4x/3)

So the integral is indeed...

(1/12)arctan(4x/3) +C

Hopefully you see where you went wrong.

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∫ 1/ (9 + 16x²) dx=
= (1/9)∫ 1/ (1 + (4x/3)²) dx=
= (1/(3*4))∫ 1/ (1 + (4x/3)²) d(4x/3)=
= (1/12)∫ 1/ (1 + (4x/3)²) d(4x/3)=
= (1/12)arctg(4x/3)+C
http://www.wolframalpha.com/input/?i=%20…

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Don't factor out 1/16. Factor out 1/12. That should end up giving you the right answer. also check and make sure you are plugging the right numbers into your inverse trig formula

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(1/12)arctg(4x/3)+C

12

keywords: problem,Calculus,integration,Calculus integration problem