Question

Consider the integral:

I= ∫ 5dx / 4 sin x + 3 cos x

(a) Use z - substitution to show that :

I = ∫ -10dz / (3z + 1)(z-3) where z = tan(x/2)

(b)  Now use the method of partial fractions to determine the integral.

Answer 1

$a)z-substitution\newline tan (\frac{x}{2})=z\newline dx=\frac{2}{1+z^{2}}dz\newline sin\ x=\frac{2z}{1+z^{2}}\newline cos\ x=\frac{1-z^{2}}{1+z^{2}}\newline I=\int \frac{5dx}{4sin\ x+3cos\ x}dx=\int \frac{5(\frac{2}{1+z^{2}})}{4(\frac{2z}{1+z^{2}})+3(\frac{1-z^{2}}{1+z^{2}})}dx\newline =\int \frac{\frac{10}{1+z^{2}}}{\frac{8z}{1+z^{2}}+\frac{3-3z^{2}}{1+z^{2}}}dz=\int \frac{\frac{10}{1+z^{2}}}{\frac{8z+3-3z^{2}}{1+z^{2}}}dz\newline =\int \frac{10}{8z+3-3z^{2}}dz=\int \frac{10}{(3z+1)(-z+3)}dz\newline =\int \frac{-10}{(3z+1)(z-3)}dz.\newline b)\newline \int \frac{-10}{(3z+1)(z-3)}dz=-[\int \frac{A}{3z+1}dz+\int \frac{B}{z-3}dz]\newline \frac{10}{(3z+1)(z-3)}=\frac{A}{3z+1}+\frac{B}{z-3}\newline multiply\ through\ by\ (3z+1)(z-3)\newline 10=A(z-3)+B(3z+1)\newline putting\ z=3\newline 10=B(3*3+1)\newline 10=10B\newline B=1\newline putting\ z=0\newline 10=-3A+B\newline 10=-3A+1\newline 9=-3A\newline A=-3\newline the\ integral\ becomes:\newline -[\int \frac{-3}{3z+1}dz+\int \frac{1}{z-3}dz]\newline = 3\frac{ln|3z+1|}{3}-ln|z-3|+C\newline =ln|3z+1|-ln|z-3|+C\newline =ln|\frac{3z+1}{z-3}|+C$