Question

How do you integrate `int sec^2x/(4-tan^2x)^(3/2)` by trigonometric substitution?

Answer 1

`tanx/(4sqrt(4-tan^2x))+C`

Explanation:

We have:

`I=intsec^2x/(4-tan^2x)^(3/2)dx`

We will first use the non-trigonometric substitution `u=tanx`, implying that `du=sec^2xdx`:

`I=int(du)/(4-u^2)^(3/2)`

Now we will apply the trigonometric substitution `u=2sintheta`. Recall that this implies that `du=2costhetad theta`.

`I=int(2costhetad theta)/(4-4sin^2theta)^(3/2)`

`=int(2costhetad theta)/(4^(3/2)(1-sin^2theta)^(3/2))`

Note that `1-sin^2theta=cos^2theta`:

`I=int(2costhetad theta)/(8(cos^2theta)^(3/2))`

`=int(costhetad theta)/(4cos^3theta)`

`=1/4int(d theta)/cos^2theta`

`=1/4intsec^2thetad theta`

`=1/4tantheta+C`

Recall that `u=2sintheta`, so `theta=arcsin(u/2)`.

`I=1/4tan(arcsin(u/2))+C`

We can simplify this: draw the triangle where sine is `u/2`, that is, the opposite side is `u` and the hypotenuse is `2`. Through the Pythagorean theorem, we see that the adjacent side is `sqrt(4-u^2)`.

Thus, the tangent is opposite over adjacent, or:

`I=1/4(u/sqrt(4-u^2))+C`

Now, since `u=tanx`:

`I=tanx/(4sqrt(4-tan^2x))+C`