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What is tan^(3/2) theta * sec(theta) d(theta) using integration?

0 votes
What is tan^(3/2) theta * sec(theta) d(theta) using integration
(with solution)
thanks!
asked Feb 21, 2013 in CALCULUS by andrew Scholar

2 Answers

+1 vote

This process to be continue infinite times

answered May 20, 2013 by richardson Scholar
Above answer is wrong.
integral sqrt of (x^2 - a^2 )dx formula is wrong
–1 vote

ʃtan^3/2(θ)×sec(θ)d(θ) = ʃtan^1/2(θ)×tan(θ)×sec(θ)d(θ)

Put sec(θ) = t

Differenciate each side by with respective t

sec(θ)tan(θ)d(θ) = dt

But tan^2(θ) = sec^2(θ) - 1 = t^2 - 1

tan(θ) = √(t^2 - 1)

Substitute tan(θ) = √(t^2 - 1) and sec(θ)tan(θ)d(θ) = dt in the above integration

                                                           = ʃ√(t^2 - 1)^1/2 dt

                                                          =ʃ(t^2 - 1)^1/4 dt

                                                          =ʃ(t^2 - 1)^1/4×1 dt

Integral product of two functions formula ʃuv = uʃv - ʃ u' ʃv

Substitute u = (t^2 - 1)^1/2 , v = 1

                                                          = (t^2 - 1)^1/4×t - ʃ1 / 4(t^2 - 1)^-3 / 4(2t - 0)tdt

                                                          =(t^2 - 1)^1/4×t - ʃ1 / 4(t^2 - 1)^-3 / 4(2t)tdt

                                                          =(t^2 - 1)^1/4×t -1/2 ʃ(t^2 - 1)^-3 / 4(t^2)dt

                                                          =(t^2 - 1)^1/4×t -1/2[(t^2 - 1)^-3/4×t^3 / 3 -ʃ-3/4(t^2 - 1)^-7/4(2t)(t^2)dt

                                                          =(t^2 - 1)^1/4×t -1/2[(t^2 - 1)^-3/4×t^3 / 3+3/2ʃ(t^2 - 1)^-7/4(t^3)dt]

Substitute t = secθ and t^2 - 1 = tan^2θ

                                                          =tan^1/2(θ) - 1/ 2[tan^-3/2(θ) sec^3(θ) / 3 + 3/2ʃ(tan)^-7/2(θ)sec(θ)tan(θ)d(θ)

                                                         =tan^1/2(θ) - 1/ 2[tan^-3/2(θ) sec^3(θ) / 3 + 3/2ʃ(tan)^-5/2(θ)sec(θ)d(θ).

This process to be continue infinite times.
                                                        

answered May 24, 2013 by diane Scholar

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