How do you integrate int 1/sqrt(3x-12sqrtx+53) using trigonometric substitution?

1 Answer
Jul 1, 2018

int1/sqrt(3x-12sqrtx+53)dx

=(2sqrt41)/(3sin(cot^(-1)(sqrt(3/41)(sqrtx-2))))+2/sqrt3ln(|csc(cot^(-1)(sqrt(3/41)(sqrtx-2)))+sqrt(3/41)(sqrtx-2)|)+C C in RR

See explanations below.

Explanation:

int1/sqrt(3x-12sqrtx+53)dx

=1/sqrt3int1/sqrt(x-4sqrtx+53/3)dx
This form of integral doesn't seem to be know: let substitute:
Let
u=sqrtx

u²=x

dx=2udu
So:
1/sqrt3int1/sqrt(x-4sqrtx+53/3)dx=1/sqrt3int(2u)/sqrt(u²-4u+53/3)du

=2/sqrt3intu/sqrt(u²-4u+53/3)du

Now we need to complete the square :

=2/sqrt3intu/sqrt((u-2)²+41/3)du

Now, let : u-2=sqrt(41/3)cot(theta)
du=-sqrt(41/3)csc(theta)²d theta

So :

2/sqrt3int(u*du)/sqrt(u²-4u+53/3)

=2/sqrt3int((sqrt(41/3)cot(theta)+2)(-sqrt(41/3)csc(theta)²))/sqrt(41/3cot(theta)²+41/3)d theta

=-2/sqrt3int((sqrt(41/3)cot(theta)+2)(cancel(sqrt(41/3))csc(theta)²))/(cancel(sqrt(41/3))sqrt(cot(theta)²+1))d theta

Because cot(theta)²+1=csc(theta)²,

-2/sqrt3int((sqrt(41/3)cot(theta)+2)csc(theta)²)/(sqrt(cot(theta)²+1))d theta=-2/sqrt3int((sqrt(41/3)cot(theta)+2)csc(theta)^(cancel(2)))/cancel(csc(theta))d theta

=-2/sqrt3intsqrt(41/3)cot(theta)csc(theta)d theta-2/sqrt3intcsc(theta)d theta

=-2/3sqrt41intcos(theta)/(sin(theta)²)d theta-2/sqrt3intcsc(theta)d theta

Now we've got two easier integrals : I won't come back on intcsc(theta)=-ln(csc(theta)+cot(theta)), you can see there the demonstration.

For the second one, remarkably, (-cos(x))^'=sin(x)
So: -intcos(theta)/(sin(theta)²)d theta=-int(f^')/(f²)=1/f=1/sin(theta)

So: -2/3sqrt41intcos(theta)/(sin(theta)²)d theta-2/sqrt3intcsc(theta)d theta

=(2sqrt41)/(3sin(theta))+2/sqrt3ln(csc(theta)+cot(theta))

But theta=cot^(-1)(sqrt(3/41)(u-2))

So:

(2sqrt41)/(3sin(theta))+2/sqrt3ln(csc(theta)+cot(theta))

=(2sqrt41)/(3sin(cot^(-1)(sqrt(3/41)(u-2))))+2/sqrt3ln(|csc(cot^(-1)(sqrt(3/41)(u-2)))+sqrt(3/41)(u-2)|)

Finally, u=sqrtx

So:

int1/sqrt(3x-12sqrtx+53)dx

=(2sqrt41)/(3sin(cot^(-1)(sqrt(3/41)(sqrtx-2))))+2/sqrt3ln(|csc(cot^(-1)(sqrt(3/41)(sqrtx-2)))+sqrt(3/41)(sqrtx-2)|)+C C in RR

\0/ Here's our answer !