Two distinct answers to same integral? I am wonder if any integral has 2 answers which might not be equivalent.
As far as I know the integral:
∫sec²x tanx dx 
has 2 answers using substitution:
When tanx=t
sex²xdx=dt
=∫tdt
=t²/2 + C
=tan²x/2 + C
When secx=t
=secxtanxdx=dt
=∫tdt
=t²/2 + C
=sec²x/2 + C
If both answers above are equivalent then please show it to me how?

The $C$ constant term in $ \frac{\tan^{2}x}{2} + C $ and $ \frac{\sec^{2}x}{2} + C $ need not be the same number, since these are solutions of an indefinite integral which is geometrically a family of curves - and they may refer to same or different curves depending on the value of their $C$s. It you are wondering how to arrive from one to the other, here is how you can do it:
$$ \frac{\tan^2x}{2} + C  = \frac{1}{2}.\frac{\sin^2x}{\cos^2x} + C $$
$$ \frac{\tan^2x}{2} + C = \frac{1}{2}.\frac{\sin^2x}{\cos^2x} +\frac{1}{2} -\frac{1}{2} + C $$
$$ \frac{\tan^2x}{2} + C = \frac{1}{2}.\Big(\frac{\sin^2x}{\cos^2x} +1\Big) -\frac{1}{2} + C $$
$$ \frac{\tan^2x}{2} + C = \frac{1}{2}.\Big(\frac{\sin^2x + \cos^2x}{\cos^2x}\Big) -\frac{1}{2} + C $$
and since, $\sin^2{\theta}+\cos^2{\theta}=1$, hence:
$$ \frac{\tan^2x}{2} + C = \frac{1}{2}.\Big(\frac{1}{\cos^2x}\Big) -\frac{1}{2} + C $$
$$ \frac{\tan^2x}{2} + C = \frac{\sec^2x}{2} -\frac{1}{2} + C $$
and $-\frac{1}{2}+C$ can be another constant, say $C_{2}$.
$$ \frac{\tan^2x}{2} + C = \frac{\sec^2x}{2} + C_{2} $$
So answering your original question - are they equivalent, YES they are!
