# Insights Solve Integrals Involving Tangent and Secant with This One Trick - Comments

Tags:
1. Mar 6, 2017

### stevendaryl

Staff Emeritus
2. Mar 6, 2017

### Mastermind01

Fascinating!

The most commonly given proof in high school for integrating sec x is multiplying the numerator and denominator by sec x + tan x. To me it seemed like a highly unintuitive step. Integrating sec x by using partial fractions - a longer but more intuitive proof, originally discovered by Barrow is the one I prefer.

This is pretty cool too. Thanks for the insight!

3. Mar 6, 2017

### vanhees71

I'd like to give this Insight 5 stars, but it just doesn't let me, because I'm not logged in for some reason when clicking at this specific Insight article. Anyway, it's a great trick, I hope to remember when I need it.

4. Mar 7, 2017

### vanhees71

Finally I could give you your 5 stars! :-).

An alternative way to integrate $\sec x$ is to use the standard subsitution
$$u=\tan \left (\frac{x}{2} \right), \quad \mathrm{d} x=\mathrm{d} u \frac{1}{1+u^2}.$$
The integrand is
$$\sec x= \frac{1}{\cos x}=\frac{1}{2 \cos^2 (x/2)-1}=\frac{1-u^2}{1+u^2}.$$
Thus you get
$$\int \mathrm{d} x \sec x= \int \mathrm{d} u \frac{2}{1-u^2} = \int \mathrm{d} u \left (\frac{1}{u+1}-\frac{1}{u-1} \right) = \ln \left |\frac{u+1}{u-1} \right|+C$$
or finaly resubstituting
$$\int \mathrm{d} x \sec x=\ln \left |\frac{\tan(x/2)+1}{\tan(x/2)-1} \right|+C.$$

5. Mar 7, 2017

### Mastermind01

@vanhees71 Awesome, I hadn't seen this one.

I still can't give five stars though.

6. Mar 7, 2017