Answered: "Solve the equation:∫2xtan9(x2)sec⁡(x2)dx"

Tagiuraoob

Answered question

2022-02-28

Solve the equation:

\int 2^{x} \tan^{9} (x^{2}) \sec (x^{2}) d x

Asa Buck

Beginner2022-03-01Added 8 answers

\int 2 x \tan^{9} (x^{2}) \sec (x^{2}) d x

We first start by simplifying the argument in the tangent and secant terms. Let's make a u-substitution where

u = x^{2} \Rightarrow d u = 2 x d x

. So we have

\int \tan^{9} (u) \sec (u) d u

The appropriate substitution is

v = \sec (u) \Rightarrow d v = \sec (u) \tan (u) d u

and we're left with

\int \tan^{9} (u) \sec (u) d u

= \int \tan (u) \tan^{8} (u) \sec (u) d u

= \int \tan^{8} (u) \sec (u) \tan (u) d u

= \int {[\tan^{2} (u)]}^{4} \sec (u) \tan (u) d u

= \int {[\sec^{2} (u) - 1]}^{4} \sec (u) \tan (u) d u

= \int {[v^{2} - 1]}^{4} d v

At this point it's just expanding into a polynomial, which is always welcome when it comes to integration:

\int {[v^{2} - 1]}^{4} d v

= \int {[(v^{2} - 1) (v^{2} - 1)]}^{2} d v

= \int {[v^{4} - 2 v^{2} + 1]}^{2} d v

= \int (v^{4} - 2 v^{2} + 1) (v^{4} - 2 v^{2} + 1) d v

= \int (v^{8} - 2 v^{6} + v^{4} - 2 v^{6} + 4 v^{4} - 2 v^{2} + v^{4} - 2 v^{2} + 1) d v

= \frac{1}{9} v^{9} - \frac{4}{7} v^{7} + \frac{6}{5} v^{5} - \frac{4}{3} v^{3} + v + C

But we know

v = \sec (u)

, and that

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