# Thread: [SOLVED] Verify trig substitution

1. ## [SOLVED] Verify trig substitution

Hey y'all!

Here is my integral:

$75 \int sec^2x tanx dx$

The book does the substitution with $u = tanx$, $du = sec^2x dx$

I worked mine out before I looked at the answer, I changed everything to terms of sin and cos and worked it that way.

$\frac{75}{2cos^2x} + C$

$\frac{75}{2} tan^2x + C$

2. Nope you're perfectly fine.

To see why, use the fact that: $\sec^2 x = \tan^2 x + 1$

\begin{aligned} \frac{75}{2\cos^2 x} + C & = \frac{75}{2}\sec^2 x + C \\ & = \frac{75}{2} \left(\tan^2 x + 1\right) + C \\ & = \frac{75}{2} \tan^2 x + {\color{red}\frac{75}{2} + C} \end{aligned}

The stuff in red is nothing more than a constant, so call it $K$

So your expression is equal to: $\frac{75}{2}\tan^2 x + K$

which is exactly what the book had.

Remember, all antiderivatives of a given function differ only by their constants which is what we've shown here.