Trig. Integral

por **stuart clark** » Ter Jul 08, 2014 04:04

Evaluation of $\displaystyle \int \sqrt[4]{\tan x}dx$

por **e8group** » Ter Jul 08, 2014 14:10

Every positive number can be written as a positive number to the power 4 . In order to us evaluate the integral ,let $0 \leq tan(x) := u^4 , u > 0 (*)$ . Taking a derivative from both sides of (*)

, we get

$sec^2(x) dx = 4 \cdot u^3 du (**)$ . Now , we use an identity to give an expression a more convenient form

$tan^2(\theta) + 1 = sec^2(\theta) , \forall \theta \in\mathbb{R}$ .Thus

$(1+ \underbrace{tan^2x}_{u^8} )dx = 4 \cdot u^3 du$ which yields

$dx = 4\frac{u^3}{1+u^8} du$ . And finally we have ,

$\int \sqrt[4]{tan(x)} dx = 4 \int \frac{u^4}{1+u^8} du$ .

I'm not sure if i'm on the right track ... Perhaps , we can attempt to use partial
fraction decomposition to write the latter integrand as a sum of fractions .

There's a trick to express u^8 + 1

as a
product of two irreducible polynomials ...

$1+u^8 = 1+ (u^4)^2 = 1^2 + 2 \cdot u^4 + (u^4)^2 - 2 \cdot u^4 = (1+u^4)^2 - (\sqrt{2}u^2)^2 = (1+u^4 + \sqrt{2} \cdot u^2 +1 ) (1+u^4 - \sqrt{2} \cdot u^2 +1 )$

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