Indefinite integrals

Let x = a tan t
dx = a sec^2 t dt
sqrt (a^2 + x^2) = sqrt (a^2 + a^2 tan^2 t) = a sqrt (1 + tan^2 t) = a sec t
so ∫ sqrt (a^2 + x^2) dx = a^2 ∫ sec^3 t dt
Then using integration by part:
v = sec t, so dv = sec t tan t dt
du = sec^2 t dt, so u = ∫ sec^2 t dt = tan t
Thus ∫ sec^3 dt = sec t tan t - ∫ tan^2 t sec t dt
= sec t tan t - ∫(sec^2 t - 1) sec t dt
= sec t tan t - ∫ sec^3 t dt + ∫ sec t dt
so 2 ∫ sec^3 t dt = sec t tan t + ∫sec t dt = sec t tan t + ln | sec t + tan t|
so ∫sec^3 t dt = (sec t tan t)/2 + (1/2) ln | sec t + tan x | + C
∫ sqrt (a^2 + x^2) dx = (x/2) sqrt (a^2 + x^2) + (a^2/2) ln | sqrt ( a^2 + x^2)/a + x/a| + C

圖片參考：http://i1099.photobucket.com/albums/g395/jasoncube/672A547D540D-5_zpsb6404f59.png


http://i1099.photobucket.com/albums/g395/jasoncube/672A547D540D-5_zpsb6404f59.png

For reference: Hyperbolic Functions
http://en.wikipedia.org/wiki/Hyperbolic_function