Calculate the integrals by partial fractions and using the indicated substitution. Show the results you get are the same.
dx/1-x^2; substitution x= sin pheta
I understand how to do the partial fraction part, but not the second part and I don't know how they are similar. Any help would be appreciated on what to do
They want you to do it two ways. The first is to change it to
dx/[2(1+x)] + dx/[2(1-x)]
which integrates to
(1/2)[ln(1+x) - ln(1-x)]
= (1/2)ln[(1+x)/(1-x)]
In the substitution method, with x = sin u
dx = cos u du
Integral dx/(1-x^2)= cos u du/1-sin^2u
= Integral du/cos u = Integral (sec u)
= (1/2)log[(1+sinu)/(1-sinu)]
= (1/2)log[(1+x)/(1-x)]
To calculate the integral using the given substitution, x = sin θ, we need to express dx in terms of dθ.
Using the derivative of the substitution, dx = cos θ dθ.
Now, let's rewrite the integral with the new variable θ and dx in terms of dθ:
∫ dx/(1 - x^2)
= ∫ cos θ dθ/(1 - sin^2θ)
= ∫ cos θ dθ/cos^2θ
By canceling out the common factor of cos θ, we get:
∫ dθ/cos θ
Now, this new integral can be evaluated easily as it has a known integral. The integral of sec θ with respect to θ is ln|sec θ + tan θ| + C, where C is the constant of integration.
Therefore, the integral of dx/(1 - x^2) using the substitution x = sin θ is ln|sec θ + tan θ| + C.
To show that the results using partial fractions and the substitution are the same, we can rewrite the integral using partial fractions:
∫ dx/(1 - x^2)
= ∫ A/(1 - x) + B/(1 + x) dx
(where A and B are constants to be determined)
Multiplying through by (1 - x)(1 + x), we have:
1 = A(1 + x) + B(1 - x)
Expanding and collecting terms, we get:
1 = (A + B) + (A - B)x
By comparing coefficients, we find A + B = 0 and A - B = 1. Solving this system of equations, we obtain A = 1/2 and B = -1/2.
Therefore, the integral using partial fractions is:
∫ dx/(1 - x^2) = (1/2)∫ dx/(1 - x) - (1/2)∫ dx/(1 + x)
= (1/2) ln|1 - x| - (1/2) ln|1 + x| + C
= (1/2) ln|1 + x| - (1/2) ln|1 - x| + C
Since ln|a| - ln|b| can be rewritten as ln|a/b|, we have:
∫ dx/(1 - x^2) = ln|1 + x| - ln|1 - x| + C
Comparing this with the result obtained from the substitution, we can see that:
ln|sec θ + tan θ| = ln|1 + x|
ln|sec θ - tan θ| = ln|1 - x|
Hence, both results are the same:
ln|1 + x| - ln|1 - x| = ln|sec θ + tan θ| + C.
By demonstrating the equivalence of the two approaches, we can conclude that they yield the same result.