Question

What is the Integral of `tan (pi/4)y dy`?

Answer 1

Assuming no errors in the question:

`I=inttan(pi/4)ydy`

Since `tan(pi/4)=1`, the integral simplifies to:

`I=intydy`

`I=inty^1dy`

Which can be integrated using the rule: `inty^ndy=y^(n+1)/(n+1)+C`

`I=y^(1+1)/(1+1)+C`

`I=y^2/2+C`

Answer 2

Assuming this specific error in the question:

`I=inttan(pi/4y)dy`

We will use substitution. First let `s=pi/4y`. Differentiating both sides yields `ds=pi/4dy`. Thus, `dy=4/pids`.

`I=inttan(s)(4/pids)`

`I=4/piinttan(s)ds`

You may already know how to integrate tangent, but this is a reminder. Rewrite tangent using sine and cosine:

`I=4/piintsin(s)/cos(s)ds`

Now, let `t=cos(s)`, implying that `dt=-sin(s)ds`. In this case, we see that `sin(s)ds=-dt`.

`I=4/piint(-dt)/t`

`I=-4/piintdt/t`

This is an important integral: `intdt/t=lnabst+C`. Thus

`I=-4/pilnabst+C`

Since `t=cos(s)`:

`I=-4/pilnabscos(s)+C`

Since `s=pi/4y`:

`I=-4/pilnabscos(pi/4y)+C`