Answer to Question #186246 in Calculus for Phyroe

Question

Integration by Partial Fractions

1.) ∫dy/(5 secy+4)

2.) ∫dx/(3-5Sin2x) from 0 to π/4 

Answer 1

$1) \int\frac{dy}{5secy+4)}\newline from trigonometry, secy=\frac{tan^2(\frac{y}{2})}{tan(\frac{y}{2})+1}\newline therefore,\newline \int\frac{dy}{(5\frac{tan^2(\frac{y}{2})}{tan(\frac{y}{2})+1}+4)}\newline \text{now, let}u=tan(\frac{y}{2}) \implies y=2tan^{-1}(u)\implies dy=(\frac{2}{1+u^2})du\newline \text{then, by substitute and arrange the terms}\newline \int\frac{dy}{5secy+4)}=\int \frac{-2(u^2-1)}{(u^2+1)(u^2+9)}du\newline \text{by partial fraction}\newline =-2\int((\frac{-1}{4(u^2+1)})+(\frac{5}{4(u^2+9)}))du\newline =-2(\frac{-tan^{-1}u}{4}+\frac{5}{3×4}tan^{-1}\frac{u}{3})+C\newline =(\frac{tan^{-1}u}{2}-\frac{5}{6}tan^{-1}\frac{u}{3})+C\newline =(\frac{tan^{-1}(tan(\frac{y}{2}))}{2}-\frac{5}{6}tan^{-1}(\frac{tan(\frac{y}{2})}{3}))+C\newline =\frac{y}{4}-\frac{5}{6}tan^{-1}(\frac{tan(\frac{y}{2})}{3}))+C\newline 2)\newline \int_{0}^{\frac{\pi}{4}} \frac{dx}{3-5sin2x} =-\int_{0}^{\frac{\pi}{4}} \frac{dx}{5sin(2x)-3}\newline put\space y=2x\implies dx=\frac{dy}{2}\newline =-\frac{1}{2}\int \frac{dy}{5sin(y)-3}=\newline from \space trigonometry, siny=2\frac{tan(\frac{y}{2})}{tan^2(\frac{y}{2})+1}\newline =-\frac{1}{2}\int \frac{dy}{5(2\frac{tan(\frac{y}{2})}{tan^2(\frac{y}{2})+1})-3}\newline =-\frac{1}{2}\int \frac{dy}{10(\frac{tan(\frac{y}{2})}{tan^2(\frac{y}{2})+1})-3}=\newline \text{now, let}u=tan(\frac{y}{2}) \implies y=2tan^{-1}(u)\implies dy=(\frac{2}{1+u^2})du\newline =-\frac{1}{2}\int \frac{-2}{3u^2-10u+3}du\newline =\int \frac{1}{3u^2-10u+3}du\newline =\int\frac{1}{(u-3)(3u-1)}du\newline \text{by partial fraction}\newline =\int\frac{1}{8(u-3)}-\frac{3}{8(3u-1)}du\newline =\frac{ln(u-3)}{8}-\frac{ln(3u-1)}{8}\newline =\frac{ln(tan(\frac{y}{2})-3)}{8}-\frac{ln(3tan(\frac{y}{2})-1)}{8}\newline =\frac{ln(tan(\frac{2x}{2})-3)}{8}-\frac{ln(3tan(\frac{2x}{2})-1)}{8}\newline =\frac{ln(tan(x)-3)}{8}-\frac{ln(3tan(x)-1)}{8}\newline Therefore,\newline \int_{0}^{\frac{\pi}{4}} \frac{dx}{3-5sin2x}=[\frac{ln(tan(x)-3)}{8}-\frac{ln(3tan(x)-1)}{8}]_{0}^{\frac{\pi}{4}}\newline =-\frac{ln(3)}{8}$