How do you integrate int 1/(x^2sqrt(x^2-36)) by trigonometric substitution?

1 Answer
Apr 6, 2018

The integral is sqrt(x^2-36)/(36x)+C.

Explanation:

Let x=6sectheta. This means that:

dx=6secthetatantheta d theta

Performing the substitution:

color(white)=int 1/(x^2sqrt(x^2-36))dx

=int1/((6sectheta)^2sqrt((6sectheta)^2-36))6secthetatantheta d theta

=int1/((6sectheta)^color(red)cancelcolor(black)2sqrt((6sectheta)^2-36))color(red)cancelcolor(black)(6sectheta)tantheta d theta

=int1/(6secthetasqrt((6sectheta)^2-36))tantheta d theta

=int1/(6secthetasqrt(36sec^2theta-36))tantheta d theta

=int1/(6secthetasqrt(36(sec^2theta-1)))tantheta d theta

=int1/(6sectheta*6sqrt(sec^2theta-1))tantheta d theta

By the Pythagorean identity tan^2theta+1=sec^2theta:

=int1/(36sectheta*sqrt(tan^2theta+1-1))tantheta d theta

=int1/(36sectheta*sqrt(tan^2theta))tantheta d theta

=int1/(36sectheta*tantheta)tantheta d theta

=int1/(36sectheta*color(red)cancelcolor(black)tantheta)color(red)cancelcolor(black)tantheta d theta

=int1/(36sectheta) d theta

=1/36int1/sectheta d theta

=1/36int1/(1/costheta) d theta

=1/36intcostheta d theta

=1/36sintheta+C

Now we want sintheta in terms of x. By our substitution earlier, we know:

6sectheta=x

sectheta=x/6

1/costheta=x/6

costheta=6/x

cos^2theta=36/x^2

-cos^2theta=-36/x^2

-cos^2theta+1=-36/x^2+1

1-cos^2theta=1-36/x^2

1-cos^2theta=(x^2-36)/x^2

sin^2theta=(x^2-36)/x^2

sintheta=sqrt((x^2-36)/x^2)

sintheta=sqrt(x^2-36)/x

Plugging this back into the integral:

color(white)=1/36sintheta+C

=1/36*sqrt(x^2-36)/x+C

=sqrt(x^2-36)/(36x)+C

That's the whole integral. Hope this helped!