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

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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)|=`

`sin^2xtanx-int(2sinxcosxsinx)/(cosx)dx=`

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

`sin^2xtanx-int(1-cos2x)dx=sin^2xtanx-x+intcos2xdx=`

Now we use partial integration

`|(t=2x),(dt=2dx)|=sin^2xtanx-x+1/2intcostdt=`

`sin^2xtanx-x+1/2sin2x+C=`

Now we use this `sin2x=2sinxcosx`

`sin^2xtanx-x+1/2cdot2sinxcosx+C=sin^2xtanx+cos^2xtanx-x+C`

`=tanx(sin^2x+cos^2x)-x=tanx-x+C`

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`.