The figure for this exercise is in the book and isn't reproduced…

(a) $x=$ ______ gives $\int dx/\sqrt{x^2+1}=\ln (\sec \theta +\tan \theta )$.

(b) The second integral converts this answer to $g=\ln (x+\sqrt{x^2+1})$. Check that $dg/dx=1/\sqrt{x^2+1}$.

(c) Verify that $\sinh g=\frac{1}{2}(e^g-e^{-g})=x$ so $g={\sinh }^{-1}x$.

(d) Substitute $x=\sinh g$ directly into $\int dx/\sqrt{x^2+1}$ and integrate.


In 37–42 substitute $x=\sin \theta$, $\cos \theta$, or $\tanh \theta$. After integration change back to $x$.

The figure for this exercise is in the book and isn't reproduced here.

This problem refers to a figure in its source book that isn't reproduced here.

Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §7.3 Trigonometric Substitutions, #36