definite integration x1

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2012-08-01 11:55:40 補充：
The followings have used integration by parts:

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∫ (0→π/4) x tan x sec²x dx

= ∫(0→π/4) x sec x (sec x tan x dx)

= ∫(0→π/4) x sec x d(sec x)

= ∫(0→π/4) (1/2) x (sec²x)

= [(1/2) x sec²x](0→π/4) - ∫(0→π/4) (1/2) sec²x dx

= [(1/2) x sec²x - (1/2) tan x](0→π/4)

= [(1/2) (π/4) sec²(π/4) - (1/2) tan (π/4)] - [0 - (1/2) tan 0]

= [(1/2) (π/4) (√2)² - (1/2) (1)]

= (π/4) - (1/2)

= (1/4) (π - 2)


2012-08-01 23:00:19 補充：
A typo :
It should be ∫(0→π/4) (1/2) x d(sec²x)
(A letter "d" has been missed.)

Let u = sec x

Then, sec x d(sec x)
= u du
= (1/2) du²
= (1/2) d sec²x

= ∫(0→π/4) x sec x d(sec x)
= ∫(0→π/4) x [(1/2) x d(sec²x)]
= ∫(0→π/4) (1/2) x d(sec²x)