Solving Integrals with Trigonometric Substitution and Partial Fractions

Integration problems often require us to apply various techniques to find a solution. In this article, we will explore how to solve two complex integration problems: one using trigonometric substitution and the other using partial fractions. Both methods are essential tools in a mathematician's arsenal and are highly regarded in academic and practical applications.

Trigonometric Substitution: int; dx / (x^2sqrt{x^2 - 9})

Consider the integral:

int; dx / (x^2sqrt{x^2 - 9})

We can use a trigonometric substitution to solve this. Since we have sqrt{x^2 - 9}, let's substitute:

x 3tan(θ)

Then, the differential dx becomes:

dx 3sec^2(θ) dθ

Substituting x 3tan(θ) into the integral:

Calculate x^2: x^2 3tan(θ)^2 9tan^2(θ)

Calculate sqrt{x^2 - 9}: sqrt{x^2 - 9} sqrt{9tan^2(θ) - 9} sqrt{9tan^2(θ) - 1} 3sec(θ)

Now substituting these into the integral:

int; dx / (x^2sqrt{x^2 - 9}) int; (3sec^2(θ) dθ) / (9tan^2(θ) * 3sec(θ))

Simplifying this, we get:

int; (3sec^2(θ) dθ) / (27tan^2(θ)sec(θ)) int; (sec(θ) / (9tan^2(θ))) dθ

Next, we can rewrite sec(θ) in terms of tan(θ) using the identities:

sec(θ) 1/cos(θ), tan(θ) sin(θ) / cos(θ)

Thus, (sec(θ) / tan^2(θ)) 1/cos(θ) * cos^2(θ)/sin^2(θ) cos(θ)/sin^2(θ)

This allows us to rewrite the integral as:

int; (cos(θ) / sin^2(θ)) dθ int; cot(θ) csch(θ) dθ

The integral of cot(θ) csch(θ) is:

-csch(θ) C

Now we need to convert back to x. Recall that:

sin(θ) 3 / sqrt(x^2 - 9), csch(θ) sqrt(x^2 - 9) / 3

Therefore, -csch(θ) -sqrt(x^2 - 9) / 3

Finally, the solution to the integral is:

int; dx / (x^2sqrt{x^2 - 9}) -sqrt(x^2 - 9) / 3 C

Partial Fractions: ∫1 / (x^2sqrt(x^2 - 9)) dx

Let's now solve the integral:

int;1/x^2sqrt(x^2 - 9) dx

We can use the substitution u x^2 - 9. Then, du/dx 2x and dx du/2x.

Substituting these into the integral, we get:

int;1/x^2sqrt(x^2 - 9) dx int; (1/2u^1/2u-9) du

Let's decompose 1/2u^1/2u-9 using partial fractions:

1/2u^1/2u-9 A/u^1/2 B/u-9

Multiplying both sides by the common denominator and simplifying, we get:

1 Au - 9 Bu^1/2

Substituting u 9 we get:

1 -9A A -1/9

Substituting u 0 we get:

1 3B√3 B 1/3√3

Substituting these values into the integral we get:

int; 1/x^2sqrt(x^2 - 9) dx -1/9 int;du/u^1/2 1/3√3 int;du/u-9

Using the power rule of integration, we can evaluate these integrals as:

-1/9 int;du/u^1/2 -2/u^1/2 C1

1/3√3 int;du/u-9 1/3√3 lnu-9 C2

Substituting back u x^2 - 9 we get:

-1/9 int;du/u^1/2 -2/(x^2 - 9)^1/2 C1

1/3√3 int;du/u-9 1/3√3 ln(x^2 - 9) C2

Therefore, the final solution to the integral is:

int;1/x^2sqrt(x^2 - 9) dx -2/(x^2 - 9)^1/2 1/3√3 ln(x^2 - 9) C where C C1 C2

Both methods, trigonometric substitution and partial fractions, are powerful tools in solving complex integrals. Understanding and practicing these techniques will significantly enhance your problem-solving capabilities in calculus.