Can you prove integral?

int1/(sqrt(x^2+1))dx=ln|x+sqrt(x^2+1)|+c

1 Answer
Mar 1, 2018

See below.

Explanation:

When dealing with integrals involving a root in the form sqrt(x^2+a^2) where a is a constant, we can make the following trigonometric substitution:

x=atan(theta)

Here, a^2=1, a=sqrt(1)=1, x=tan(theta)

dx=sec^2(theta)d theta

Substituting yields:

intsec^2(theta)/sqrt(tan^2(theta)+1)d theta

Recall the identity

tan^2(theta)+1=sec^2(theta). Apply it to what's under the root:

intsec^2(theta)/sqrt(sec^2(theta))d theta

sqrt(sec^2(theta))=|sec(theta)|=sec(theta). We'll have to assume secant will remain positive since we're working with an indefinite integral.

intsec^(cancel(2)1)(theta)/cancel(sec(theta))d theta

intsec(theta)d theta

This is a common integral, I'll show a proof at the end, but it should be memorized:

intsec(theta)d theta=ln|sec(theta)+tan(theta)|+C

We must rewrite in terms of x. We said tan(theta)=x, it was our initial substitution.

To find sec(theta), recall the identity

1+tan^2(theta)=sec^2(theta)
Substitute in x=tan(theta) meaning that tan^2(theta)=x^2

sec^2(theta)=1+x^2

Take the root of both sides:

sec(theta)=sqrt(1+x^2)

Thus

intsec(theta)d theta=intdx/sqrt(x^2+1)=ln|sqrt(1+x^2)+x|+C

Proof of intsecthetad theta=ln|sec(theta)+tan(theta)|+C:

intsecthetad theta=intsectheta (sectheta+tantheta)/(sectheta+tantheta)

Multiply the integrand by (sectheta+tantheta)/(sectheta+tantheta). This is the same as multiplying by 1.

Make the following substitution:

u=sectheta+tantheta

du=(sec(theta)tan(theta)+sec^2(theta))d theta

If we multiply out the numerator of our integrand, we have

intsectheta(sectheta+tantheta)/(sectheta+tantheta)=int(sec^2theta+secthetatantheta)/(sectheta+tantheta)

So, du appears in the numerator and u is the denominator. . We now have

int(du)/u=ln|u|+C

Rewrite in terms of theta:

intsec(theta)=ln|sec(theta)+tan(theta)|+C