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 !