Fairly Basic Integration?

Question

Hi, can anyone help me can't remember how to integrate x/(x+1) and 1/sin^2(x). Thanks.

P.S Obvioulsy with respect to x for anyone who is really pedantic :-)

Answer 1

the 2 problems can be done more easily

1) x/(x+1)


because numerator and denominator have same power of x we can reduce the ratio

x/x+1 = (x+1-1)/(x+1) = 1 - 1/(x+1)

integral of 1 = x
interal of 1/(x+1) = ln x
so integral = x - ln x +C some arbritary constant

2) 1/sin^2 x = cosec^2 x

integral = - cot x + C based on standard formula

i

Answer 2

The first one is pretty straighforward. Since it's a fraction, try to head for a du/u form:

x/(x+1)*dx = ((x+1) - 1)/(x+1)*dx = (1 - 1/(x+1))*dx = dx - du/u

Where u = x+1

Integrating: x - ln(u) = x - ln(x+1)

The second one is more of a pain. Again begin with some rearrangement:

dx/sin(x)^2 = (cos(x)^2 + sin(x)^2)/sin(x)^2*dx = (cos(x)/sin(x))^2*dx + dx

Of couse the dx integrates to x. Work on the other term and integrate by parts:

(cos(x)/sin(x))^2*dx = u*dv

cos(x) = u, so du = - sin(x)*dx

cos(x)*dx/sin(x)^2 = dv

This has the form dz/z^2 where z = sin(x)

Integrating: v = -1/z = -1/sin(x)

Put together the integration by parts so far:

Int((cos(x)/sin(x))^2*dx) = Int(u*dv) = u*v - Int(v*du) = -cos(x)/sin(x) - Int((-sin(x)*dx) * (-1/sin(x))) = -cos(x)/sin(x) - Int(dx) = -cos(x)/sin(x) - x

Combine this with the integration of dx from above and:

Int(dx/sin(x)^2) = -cos(x)/sin(x)

Answer 3

Wouldn't the second one be easier if one wrote it
as csc^2(x)? The integral of this is -cot(x). Done!

Answer 4

integration by parts

x - log(1 + x)

Second one should be -cosx/sinx (i think)