Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've began a course on PDE's; and I am very rusty with integration.

One question required the integral $\displaystyle\int\dfrac{1}{1+x^2}dx$ . My first instinct was to substitute $x=\sinh(z)\implies dx=\cosh(z)~dz$ and we get the integral $$\displaystyle\int\dfrac{\cosh(z)}{1+\sinh^2(z)}dz=\displaystyle\int\dfrac{1}{\cosh(z)}dz$$

at which point I didn't really know how to evaluate it any further.. Tried a different route and $x=\tan(z)$ did much better, and got the result $\arctan(x)$.

But, that is definitely not the result of the last integral of my first approach, and I cannot see any mistakes I made in getting to it; what have I done?

share|improve this question
2  
Differentiate $\arctan (\sinh z)$. You get $\dfrac{1}{\cosh z}$ indeed. –  Daniel Fischer Jan 16 at 18:14

1 Answer 1

up vote 5 down vote accepted

I think that trigonometric substitution is the most straightforward approach here.

Given your integral, we put $x =\tan \theta \implies dx = \sec^2 \theta \,d\theta, \quad \theta = \arctan x$

Substituting correctly, you should obtain, in the end,

$$\int \frac{1}{1 + x^2}\,dx = \int \frac{\sec^\theta\,d\theta}{1 + \tan^2 \theta} = \int \frac {\sec^2 \theta}{\sec^2\theta}\,d\theta = \int d\theta = \theta + C = \arctan(x) + C$$


If we substitute, instead, $x = \sinh z$,

$$\displaystyle\int\dfrac{\cosh(z)}{1+\sinh^2(z)}dz=\displaystyle\int\dfrac{1}{\cosh(z)}dz = \arctan(\sinh z) + C=\arctan(x) + C$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.