Question

evaluate the integral from 0 to pie/8. (2x+sec2xtan2x)dx

Answer 1

look in the book and see what the antiderivative of
\[\sec^2(x)\tan^2(x)\] is

Answer 2

oh wait sorry. this is
\[\sec(2x)\tan(2x)\] yes?

Answer 3

sec 2x=sec2x+tan2x

Answer 4

tan2x=sec^22x

Answer 5

:(\

Answer 6

sec^2 tan^2 = (sec tan)^2

Answer 7

no no easy way
the derivative of secant is secant tangent

Answer 8

so anti-derivative of secant tangent is secant.

Answer 9

good notation does help ;)

Answer 10

your anti-derivative is just
\[x^2+\frac{1}{2}\sec(2x)\]

Answer 11

the one half because you have
\[\sec(2x)\] not \[\sec(x)\]

Answer 12

oh

Answer 13

ok

Answer 14

so now just plug in
\[\frac{\pi}{8}\] and 0 and be happy