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 147477 by vvvv last updated on 21/Jul/21

Commented by Rustambek last updated on 21/Jul/21

$$salom$$

Answered by puissant last updated on 21/Jul/21

$$={\int }_0^{\frac{\pi }{2}}\frac{cos\left(x\right)+18sin\left(x\right)}{-4cos\left(x\right)+3sin\left(x\right)}dx$$$$={[\left(\frac{1\times (-4)+3\times 18}{{(-4)}^2+3^3}\right)x+\left(\frac{1\times 3-18\times (-4)}{{(-4)}^2+3^2}\right)ln\mid -4cos\left(x\right)+3sin\left(x\right)\mid ]}_0^{\frac{\pi }{2}}$$$$={[2x+3ln\mid -4cos\left(x\right)+3sin\left(x\right)\mid ]}_0^{\frac{\pi }{2}}$$$$=(\pi +3ln3)-3ln4$$$$\Rightarrow I=\pi +3ln\left(\frac{3}{4}\right)...$$

Commented by vvvv last updated on 21/Jul/21

$$\mathit{y}\mathit{e}\mathit{s}\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{i}\mathit{s}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}\mathit{s}$$

Commented by gsk2684 last updated on 21/Jul/21

$$\int \frac{A(-c\sin x+d\cos x)+B(c\cos x+d\sin x)}{c\cos x+d\sin x}dx$$$$\int \frac{(-Ac+Bd)\sin x+(Ad+Bc)\cos x}{c\cos x+d\sin x}dx$$$$\therefore solvingdB-cA=b,cB+dA=a$$$$\Rightarrow$$$$dB-cA-b=0$$$$cB+dA-a=0$$$$\Rightarrow B=\frac{ac+bd}{c^2+d^2},A=\frac{ad-bc}{c^2+d^2}$$$$\therefore \int \left(\frac{ad-bc}{c^2+d^2}\right)\left(\frac{-c\sin x+d\cos x}{c\cos x+d\sin x}\right)dx+\int \left(\frac{ac+bd}{c^2+d^2}\right)dx$$$$\left(\frac{ad-bc}{c^2+d^2}\right)\ln (c\cos x+d\sin x)+\left(\frac{ac+bd}{c^2+d^2}\right)x$$$$$$

Commented by vvvv last updated on 21/Jul/21

$$?$$

Commented by puissant last updated on 21/Jul/21

$$\int \frac{acos\left(x\right)+bsin\left(x\right)}{ccos\left(x\right)+dsin\left(x\right)}dx$$$$=\left(\frac{ac+bd}{c^2+d^2}\right)x+\left(\frac{ad-bc}{c^2+d^2}\right)ln\mid ccos\left(x\right)+dsin\left(x\right)\mid +c$$

Commented by mathmax by abdo last updated on 21/Jul/21

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{cosx}+18\mathrm{sinx}}{4\mathrm{cosx}-3\mathrm{sinx}}\mathrm{dx}\Rightarrow \mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{1+18\mathrm{tanx}}{4-3\mathrm{tanx}}\mathrm{dx}$$$${=}_{\mathrm{tanx}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{1+18\mathrm{t}}{4-3\mathrm{t}}\times \frac{\mathrm{dt}}{1+{\mathrm{t}}^2}=-{\int }_0^{\mathrm{\infty }}\frac{18\mathrm{t}+1}{(3\mathrm{t}-4)({\mathrm{t}}^2+1)}\mathrm{dt}$$$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{t}\right)=\frac{18\mathrm{t}+1}{(3\mathrm{t}-4)({\mathrm{t}}^2+1)}$$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{a}}{3\mathrm{t}-4}+\frac{\mathrm{bt}+\mathrm{c}}{{\mathrm{t}}^2+1}$$$$\mathrm{a}=\frac{18.\frac{4}{3}+1}{\frac{16}{9}+1}=\frac{25}{25}\times 9=9$$$${\lim }_{\mathrm{t}\to \mathrm{\infty }}\mathrm{tF}\left(\mathrm{t}\right)=0=\frac{\mathrm{a}}{3}+\mathrm{b}\Rightarrow \mathrm{b}=-\frac{9}{3}=-3$$$$\mathrm{F}\left(0\right)=-\frac{1}{4}=-\frac{\mathrm{a}}{4}+\mathrm{c}=-\frac{9}{4}+\mathrm{c}\Rightarrow \mathrm{c}=\frac{9}{4}-\frac{1}{4}=2\Rightarrow$$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{9}{3\mathrm{t}-4}+\frac{-3\mathrm{t}+2}{{\mathrm{t}}^2+1}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\mathrm{F}\left(\mathrm{t}\right)\mathrm{dt}=3{\int }_0^{\mathrm{\infty }}(\frac{1}{\mathrm{t}-\frac{4}{3}}-\frac{3}{2}\times \frac{2\mathrm{t}}{{\mathrm{t}}^2+1}+\frac{2}{{\mathrm{t}}^2+1})\mathrm{dt}$$$$=3{[\log \mid \frac{\mathrm{t}-\frac{4}{3}}{\sqrt{{\mathrm{t}}^2+1}}\mid ]}_0^{\mathrm{\infty }}+6{\left[\mathrm{arctant}\right]}_0^{\mathrm{\infty }}$$$$=3(-\log \left(\frac{4}{3}\right))+6\left(\frac{\pi }{2}\right)=3\pi -3\log \left(\frac{4}{3}\right)\Rightarrow$$$$\mathrm{I}=3\log \left(\frac{4}{3}\right)-3\pi =6\log 2-3\log 3-3\pi$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com