Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 87854 by jagoll last updated on 06/Apr/20

$$\int \frac{1}{\sin \mathrm{x}+2\cos \mathrm{x}+3}\mathrm{dx}$$

Commented by mathmax by abdo last updated on 06/Apr/20

$$I=\int \frac{dx}{2cosx+sinx+3}wedothechangementtan\left(\frac{x}{2}\right)=t\Rightarrow$$$$I=\int \frac{2dt}{(1+t^2)(2\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}+3)}=\int \frac{2dt}{2-2t^2+2t+3+3t^2}$$$$=\int \frac{2dt}{5+2t+t^2}=2\int \frac{dt}{t^2+2t+5}=2\int \frac{dt}{{(t+1)}^2+4}$$$${=}_{t+1=2u}2\int \frac{2du}{4(1+u^2)}=arctan\left(\frac{t+1}{2}\right)+C$$$$=arctan(\frac{1}{2}(1+tan\left(\frac{x}{2}\right))+C$$

Commented by john santu last updated on 07/Apr/20

$$\mathrm{let}\mathrm{x}=2\mathrm{arc}\tan \mathrm{t}\Rightarrow \mathrm{dx}=\frac{2\mathrm{dt}}{1+{\mathrm{t}}^2}$$$$\int \frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}+\frac{2-{2\mathrm{t}}^2}{1+{\mathrm{t}}^2}+3)}$$$$\int \frac{2\mathrm{dt}}{{\mathrm{t}}^2+2\mathrm{t}+5}=\int \frac{2\mathrm{dt}}{{(\mathrm{t}+1)}^2+4}$$$$\frac{1}{2}\int \frac{\mathrm{d}(\mathrm{t}+1)}{1+{\left(\frac{\mathrm{t}+1}{2}\right)}^2}=\int \frac{\mathrm{d}\left(\frac{\mathrm{t}+1}{2}\right)}{1+{\left(\frac{\mathrm{t}+1}{2}\right)}^2}$$$$=\mathrm{arc}\tan \left(\frac{\mathrm{t}+1}{2}\right)+\mathrm{c}$$$$=\mathrm{arc}\tan \left(\frac{\tan \left(\frac{\mathrm{x}}{2}\right)+1}{2}\right)+\mathrm{c}$$$$$$

Commented by jagoll last updated on 06/Apr/20

$$\mathrm{waw}...\mathrm{thanks}\mathrm{all}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com