76
Math and Science / Re: Trig Substitution
« on: January 24, 2013, 11:12:25 pm »
integral(1/sqrt(16-x^2))dx

using trig substitution
sin(-)=x/4
4sin(-)=x
sqrt(4^2-x^2)=4cos(-)
dx = d/dx(x)
dx = d/dx(4sin(-))
dx = 4cos(-)d(-)
substitute to original equation:
integral((1/4cos(-))(4cos(-)d(-)))
= integral (1 d(-))
integrate it and you get (-)+C.
now since sin(-)=x/4
(-) = arcsin(x/4)
therefore (-)+C = arcsin(x/4)+C

using trig substitution
sin(-)=x/4
4sin(-)=x
sqrt(4^2-x^2)=4cos(-)
dx = d/dx(x)
dx = d/dx(4sin(-))
dx = 4cos(-)d(-)
substitute to original equation:
integral((1/4cos(-))(4cos(-)d(-)))
= integral (1 d(-))
integrate it and you get (-)+C.
now since sin(-)=x/4
(-) = arcsin(x/4)
therefore (-)+C = arcsin(x/4)+C