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

Question

Ezra Adams · Accepted Answer

#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.

#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 !

Scarlett Allison · Answer

undefined To integrate $ \int \frac{1}{\sqrt{3x - 12\sqrt{x} + 53}} $ using trigonometric substitution:

1. First, complete the square inside the square root to make it easier to work with.
2. Perform the substitution $ x = 3(u + 2)^2 $.
3. Express the integral in terms of $ u $.
4. Solve the integral using trigonometric techniques.
5. Finally, revert to the original variable $ x $ if needed.

This process helps simplify the integral and makes it amenable to trigonometric techniques. If you need further clarification or assistance with any step, feel free to ask.