Inverse Trig Integrals
int(-8)/(1+x^2)dx
8arccot(x) or -8arctan(x) (both +c)
y=sin^-1(2x+5)
dy/dx=1/sqrt(1-(2x+5)^2)*2
sin^-1(-.5)
-pi/6
derive:
y= 2^x
dy/dx=2^xln(2)
int_0^(1/2)(sin^(-1)x)/sqrt(1-x^2)dx
pi^2/72
y=arctan(cos(theta))
dy/(d theta)=(-sin(x))/(cos^2(x)+1
tan(arctan(10))
10
int (x^3+5x^2-32x-7)/(x-4)dx
x^3/3+9/2x^2+4x+9ln(abs(x-4))+c
int(1+x)/(1+x^2)dx
1/2ln(x^2+1)+tan^-1(x)+c
y=tan^-1(x^3)
dy/dx=(3x^2)/(1+x^6
sin(arctan(10))
(10sqrt(101))/101
derive:
y= 5^(x+1)
dy/dx=5^(x+1)ln(5)
int dx/(sqrt(x)*(1+x))
2tan^-1(sqrt(x)) +c
y=sin^-1(cos^-1(x))
1/sqrt(1-(cos^-1(x))^2)*-1/sqrt(1-x^2)
csc(cos^-1(3/5))
5/4
int(x/(x^4+2x^2+5))dx
1/4*tan^-1((x^2+1)/2)+c