[SOLVED] Verify trig substitution

**mollymcf2009** · February 23rd 2009, 06:10 PM

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.

My answer:

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

Book answer:

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

Is my answer wrong?

**o_O** · February 23rd 2009, 06:32 PM

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.

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