How do you solve this integral??

Question

Please help me to solve these integral using the chain rule:

the integral of 1/(1+y^2)

Answer 1

You have to use trigonmetric substitution, which is by far, the hardest type of integration. It also involves making diagrams of triangles, which I hope you will bear with me with this limited interface.

So we want to solve for Integral (1/(1+y^2))dy. You have to use trigonometric substitution, and

let y = tan(x). Then dy = (sec(x))^2, or "secant squared x", and we just directly substitute that.

Integral (1/(1+y^2))dy =
Integral (1/(1 + (tan(x))^2) (sec(x))^2 dx =

Note: In trigonometric substitution, unlike the regular type, you're also replacing dy for what you solved for.

Using the trig identity (sec(x))^2 = (tan(x))^2 + 1, we get

Integral (1/(sec(x))^2 * (sec(x))^2) dx = (the two values will cancel each other out into 1)
Integral (1)dx =

x + C

Remember that above, we let y = tan(x), so we have to solve for x, and we get x = tan(inverse)y, or tan(x) with a -1 value between the "tan" and the "x". The reason for this step is we require an answer __in terms of y__.

So, our actual answer is tan(inverse)(y) + C

Answer 2

a simpler way is that you know that (d/dy) arctan y =1/(1+y^2) = f'(x) so you solve by saying it is f(x)= arctan(y) + c

Answer 3

dunno how to solve but the answer is:
the integral of 1/(1+y^2)= arc tan ( y ) +C