# The following integrals require a preliminary step such as a change of variables

The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals.
$\int \frac{\mathrm{cos}\theta }{\left({\mathrm{sin}}^{3}\theta -4\mathrm{sin}\theta \right)}d\theta$
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Brittany Patton
We first use substitution
Let,
$z=\mathrm{sin}\theta$
$dz=\mathrm{cos}\theta d\theta$
$\int \frac{\mathrm{cos}\theta }{\left({\mathrm{sin}}^{3}\theta -4\mathrm{sin}\theta \right)}d\theta$
$=\int \frac{1}{\left({z}^{3}-4z\right)}dz$
$=\int \frac{1}{z\left({z}^{2}-4\right)}dz=\int \frac{1}{z\left(z+2\right)\left(z-2\right)}dz$
Using partial decomposition we get
$\frac{1}{z\left(z+2\right)\left(z-2\right)}=\frac{A}{z}+\frac{B}{z+2}+\frac{C}{z-2}$
$\frac{1}{z\left(z+2\right)\left(z-2\right)}=\frac{A\left(z+2\right)\left(z-2\right)+Bz\left(z-2\right)+Cz\left(z+2\right)}{z\left(z+2\right)\left(z-2\right)}$
$1=A\left(z+2\right)\left(z-2\right)+Bz\left(z-2\right)+Cz\left(z+2\right)$
$1=A{z}^{2}-4A+B{z}^{2}+C{z}^{2}+2Cz$
$1=\left(A+B+C\right){z}^{2}+\left(-2B+2C\right)z-4A$
$A+B+C=0$
$A+B+C=0$
$-2B+2C=0$
$-4A=1$
$A=-\frac{1}{4},B=\frac{1}{8},C=\frac{1}{8}$
$\int \frac{1}{z\left(z+2\right)\left(z-2\right)}dz$
$=\int \left(\frac{-\frac{1}{4}}{z}+\frac{\frac{1}{8}}{z+2}+\frac{\frac{1}{8}}{z-2}\right)dz$