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iohanetc

Answered question

2021-10-17

The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals.

\int \frac{\cos θ}{(\sin^{3} θ - 4 \sin θ)} d θ

Answer & Explanation

Brittany Patton

Skilled2021-10-18Added 100 answers

We first use substitution
Let,

z = \sin θ

d z = \cos θ d θ

\int \frac{\cos θ}{(\sin^{3} θ - 4 \sin θ)} d θ

= \int \frac{1}{(z^{3} - 4 z)} d z

= \int \frac{1}{z (z^{2} - 4)} d z = \int \frac{1}{z (z + 2) (z - 2)} d z

Using partial decomposition we get

\frac{1}{z (z + 2) (z - 2)} = \frac{A}{z} + \frac{B}{z + 2} + \frac{C}{z - 2}

\frac{1}{z (z + 2) (z - 2)} = \frac{A (z + 2) (z - 2) + B z (z - 2) + C z (z + 2)}{z (z + 2) (z - 2)}

1 = A (z + 2) (z - 2) + B z (z - 2) + C z (z + 2)

1 = A z^{2} - 4 A + B z^{2} + C z^{2} + 2 C z

1 = (A + B + C) z^{2} + (- 2 B + 2 C) z - 4 A

A + B + C = 0

A + B + C = 0

- 2 B + 2 C = 0

- 4 A = 1

A = - \frac{1}{4}, B = \frac{1}{8}, C = \frac{1}{8}

\int \frac{1}{z (z + 2) (z - 2)} d z

= \int (\frac{- \frac{1}{4}}{z} + \frac{\frac{1}{8}}{z + 2} + \frac{\frac{1}{8}}{z - 2}) d z

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