Trigonometric Substitution Integration

Question

2.) ∫ ((9+25x^2)^(1/2)*dx)/(x^2)

Answer 1

Hi Helayna!

Because we have a positive x² term and a positive constant under the square root, this suggests that we do a substitution using tangent. The constant 9 dictates how the substitution should look; I need to change the 25 into 9 when I do my substitution so that I can factor out the 9 and use the identity 1+tan²(θ)=sec²(θ). So, the right substitution is x=(3/5)tan(θ). Then x² would be (9/25)tan²(θ), which allows me to change the 25 to 9 like I want. We also find the differential dx before we change everything in the integral to θ: dx=(3/5)sec²(θ)dθ.

So, our integral ∫ ((9+25x²)^(1/2)dx)/x² becomes:

∫((9+25(9/25)tan²(θ))^(1/2)(3/5)sec²(θ)dθ)/((9/25)tan²(θ))

=(3/5)/(9/25)∫((9+9tan²(θ))^(1/2)sec²(θ))/tan²(θ) dθ

=(3/5)*(25/9)∫((9(1+tan²(θ)))^(1/2)sec²(θ))/tan²(θ) dθ

=(5/3)∫((9sec²(θ))^(1/2)sec²(θ))/tan²(θ) dθ

=3*(5/3)∫(sec(θ)sec²(θ))/tan²(θ) dθ

=5∫sec³(θ)/tan²(θ) dθ

Now for the often challenging part: how to integrate the resulting trigonometric integral. We can try to rewrite the integral using identities and look for a u-substitution. Sometimes identities don't work and we need to look at methods like integration by parts. Though this integral could be done either way, I have found integration by parts to be easier here. Let u=sec(θ); then du=sec(θ)tan(θ)dθ. Let dv=sec²(θ)/tan²(θ)dθ. If I integrate this (this is a basic u-substitution integral in which we would do u=tan(θ) because the derivative is the top, sec²(θ), and substituting turns the fraction into 1/u², whose antiderivative is -1/u), I get v=-1/tan(θ), a much simpler fraction. So,

5∫sec³(θ)/tan²(θ) dθ=5(-sec(θ)/tan(θ)-∫-sec(θ)tan(θ)/tan(θ)dθ)

=5(-sec(θ)/tan(θ)+∫sec(θ)dθ) (The tan(θ)'s in the new integral nicely cancelled out)

=5(-sec(θ)/tan(θ)+ln|sec(θ)+tan(θ)|)+C

Now we need to turn θ back into x. This can be tricky at times. Normally we draw a triangle representing our trigonometric substitution and solve for the missing side and use the sides of the triangle to replace all of the trigonometric functions we have with expressions with x. Here I don't have to mess with a triangle because my only trigonometric functions are tangent and secant and I have x related to tangent and I have an identity relating tangent to secant (since 1+tan²(θ)=sec²(θ), then sec(θ) is just (1+tan²(θ))^(1/2)--this only works because we usually assume θ is in the restricted domain (-π/2, π/2) where secant is positive). So, I have the following:

x=(3/5)tan(θ) means tan(θ)=(5/3)x,

tan²(θ) is (25/9)x², and

sec(θ)=(1+(25/9)x²)^(1/2)

We plug all of this in to get x back:

5(-(1+(25/9)x²)^(1/2)/((5/3)x)+ln|(1+(25/9)x²)^(1/2)+(5/3)x|)+C

If we distribute the 5 and flip over the (5/3), we would get this as the antiderivative:

-3(1+(25/9)x²)^(1/2)/x+5ln|(1+(25/9)x²)^(1/2)+(5/3)x|+C

The techniques I used here may not be what is ideal for every problem, and as you can see these typically are lengthy problems to do by hand. But I sincerely hope this helps you and gives you some ideas as to how to do these problems.

Answer 2

Substitution: x = 3/5tant, dx = 3/5sec²tdt. Our integral equel I = ∫(9 + 9tan²t)^1/23/5sec²t/(9/25tan²t)dt =

9/5·25/9∫sec³t/tan²tdt = 5∫dt/(costsin²t)= 5∫costdt/[(1 - sin²t)sin²t] Now substitution sit = u gives I = 5∫du/[u²(1 - u²)] . This integral you can evaluate using Partial Fractions Decomposition.