Find the integral of [csc (pi x) cot (pi x)] and [1/(e^-x +1)]?

Question

disregard the last integral

Answer 1

For the second integral:

1) Use the substitution u = e^(-x) + 1. Solve this for x and you have x = -ln(u-1), from which you can get dx. You'll need to use partial fraction decomposition.

2) Write the integrand as (1+e^(-x) - e^(-x))/(e^(-x) + 1). Split this into
(1+e^(-x))/(e^(-x) + 1) - e^(-x)/(e^(-x) + 1), that is,
1 - [e^(-x)/(e^(-x) + 1)]. Then use the substitution
u = e^(-x) + 1, so du = -e^(-x) dx.

Answer 2

csc(ax)cot(ax)=(1/sin(ax))*(cos(ax)/sin(ax)):
let u=sin(ax), then du=a*cos(ax)dx,
and the integrand becomes
a*du/u^2,
and the integral is
(a/3)*u^3 = (pi/3)*(sin^3(ax)) + C

Answer 3

1/♠ y(x)*dx = dx*(1/sin) *(cos/sin) =dx*cos/sin^2;
u=sin(pi x), du =pi*dx*cos(pi x), du/pi =dx*cos(pi x);
y(u)*du = (du/pi) /u^2;
♣ Y(u) = (-1/pi)/u = C - (1/pi)/sin(pi x);
2/♦ y(x)*dx = dx/(exp(-x) +1) = dx *exp(x) /((1+exp(x));
u=exp x, du =dx *exp x; y(u)*du = du/(1+u);
♥ Y(u) = ln(1+u)= ln(1+ e^x) +C;