Proved ∫tan^2 Ө dӨ = ∫(sec^2 Ө -1)dӨ = tan Ө- Ө + c.  please explain.

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tiburtius eNotes educator| Certified Educator

We will use the following facts:

`int(f(x)+g(x))dx=intf(x)dx+intg(x)dx`       (1)

`int sec^2xdx=int(dx)/(cos^2x)dx=tanx+ C`              (2)

`int u(x)dv(x)dx=u(x)v(x)-int v(x)du(x)` <-- partial integration

Let's first calculate `int tan^2xdx` (I will use `x` instead of `theta`).

`int tan^2xdx=int(sin^2x)/(cos^2x)dx=|(u=sin^2x, du=2sinxcosxdx),(dv=dx/(cos^2x), v=tanx)|=`


Now we use this `sin^2x=(1-cos(2x))/2`


Now we use partial integration



Now we use this `sin2x=2sinxcosx`



The last equality follows because `sin^2x+cos^2x=1`

So we have proved that `inttan^2xdx=tanx-x+C`

Let's now calculate the other integral

`int (sec^2x-1)dx=int sec^2xdx-intdx=tanx-x+C`

First equality comes from (1) and second from (2)

which proves that two integrals are equal to `tanx-x+C`.

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