Question Number 118491 by Lordose last updated on 18/Oct/20
$$ \\ $$$$...\:\blacklozenge\mathrm{Advanced}\:\mathrm{Calculus}\blacklozenge... \\ $$$$ \\ $$$$\mathrm{Evaluate}:: \\ $$$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{sec}\theta}{\:\sqrt{\mathrm{4tan}^{\mathrm{2}} \theta+\mathrm{5}}}\mathrm{d}\theta \\ $$$$ \\ $$$$...\spadesuit\boldsymbol{\mathrm{L}\phi\mathrm{rD}}\:\boldsymbol{\varnothing\mathrm{sE}}\spadesuit... \\ $$$$ \\ $$$$...\clubsuit\boldsymbol{\mathrm{GooD}}\:\boldsymbol{\mathrm{LucK}}\clubsuit \\ $$
Answered by MJS_new last updated on 18/Oct/20
![∫((sec θ)/( (√(5+4tan^2 θ))))dθ= =sign (cos x)((√5)/5)∫(dθ/( (√(1−(1/5)sin^2 θ))))= =sign (cos x)((√5)/5)F (x∣(1/5)) +C ⇒ Ω is not defined [plot ((sec θ)/( (√(5+4tan^2 θ)))) and you′ll see why]](Q118610.png)
$$\int\frac{\mathrm{sec}\:\theta}{\:\sqrt{\mathrm{5}+\mathrm{4tan}^{\mathrm{2}} \:\theta}}{d}\theta= \\ $$$$=\mathrm{sign}\:\left(\mathrm{cos}\:{x}\right)\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\int\frac{{d}\theta}{\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{sin}^{\mathrm{2}} \:\theta}}= \\ $$$$=\mathrm{sign}\:\left(\mathrm{cos}\:{x}\right)\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\mathrm{F}\:\left({x}\mid\frac{\mathrm{1}}{\mathrm{5}}\right)\:+{C} \\ $$$$\Rightarrow\:\Omega\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined} \\ $$$$\left[\mathrm{plot}\:\frac{\mathrm{sec}\:\theta}{\:\sqrt{\mathrm{5}+\mathrm{4tan}^{\mathrm{2}} \:\theta}}\:\mathrm{and}\:\mathrm{you}'\mathrm{ll}\:\mathrm{see}\:\mathrm{why}\right] \\ $$