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Question Number 59631 by Mr X pcx last updated on 12/May/19

$$1)calculate{\int }_0^{2\pi }\frac{dx}{acosx+bsinx}$$$$witha,breals$$$$2)findalso{\int }_0^{2\pi }\frac{cosxdx}{{(acosx+bsinx)}^2}and$$$${\int }_0^{2\pi }\frac{sinxdx}{{(acosx+bsinx)}^2}$$$$3)findthevalueof{\int }_0^{2\pi }\frac{dx}{2cosx+\sqrt{3}sinx}$$

Commented by Mr X pcx last updated on 12/May/19

$$4)findthevalueof{\int }_0^{2\pi }\frac{cosxdx}{{(3cosx+\sqrt{3}sinx)}^2}$$

Commented by maxmathsup by imad last updated on 15/May/19

$$wehaveI={\int }_0^{\pi }\frac{dx}{acosx+bsinx}+{\int }_{\pi }^{2\pi }\frac{dx}{acosx+bsinx}=H+K$$$$H{=}_{tan\left(\frac{x}{2}\right)=t}{\int }_0^{\mathrm{\infty }}\frac{1}{a\frac{1-t^2}{1+t^2}+b\frac{2t}{1+t^2}}\frac{2dt}{1+t^2}={\int }_0^{\mathrm{\infty }}\frac{2dt}{a-{at}^2+2bt}$$$$={\int }_0^{\mathrm{\infty }}\frac{-2dt}{{at}^2-2bt-a}rootsof{at}^2-2bt-a=0$$$${\mathrm{\Delta }}^{{}^{\prime }}=b^2+a^2>0fora\ne 0\Rightarrow t_1=\frac{b+\sqrt{a^2+b^2}}{a}$$$$t_2=\frac{b-\sqrt{a^2+b^2}}{a}\Rightarrow H={\int }_0^{\mathrm{\infty }}\frac{-2}{a(t-t_1)(t-t_2)}dt=\frac{-2}{a(t_1-t_2)}{\int }_0^{\mathrm{\infty }}(\frac{1}{t-t_1}-\frac{1}{t-t_2})dt$$$$=\frac{-2}{\frac{2a\sqrt{a^2+b^2}}{a}}{[ln\mid \frac{t-t_1}{t-t_2}\mid ]}_0^{+\mathrm{\infty }}=\frac{-1}{\sqrt{a^2+b^2}}(-ln\mid \frac{t_1}{t_2}\mid )=\frac{1}{\sqrt{a^2+b^2}}ln\mid \frac{b+\sqrt{a^2+b^2}}{b-\sqrt{a^2+b^2}}\mid$$$$\Rightarrow H=\frac{1}{\sqrt{a^2+b^2}}ln\left(\frac{\sqrt{a^2+b^2}+b}{\sqrt{a^2+b^2}-b}\right)$$$$K{=}_{x=t+\pi }{\int }_0^{\pi }\frac{dt}{-acost-bsint}=-\frac{1}{\sqrt{a^2+b^2}}ln\left(\frac{\sqrt{a^2+b^2}+b}{\sqrt{a^2+b^2}-b}\right)\Rightarrow$$$$I=0$$

Commented by maxmathsup by imad last updated on 15/May/19

$$2)letf\left(a\right)={\int }_0^{2\pi }\frac{dx}{acosx+bsinx}\Rightarrow f^{{}^{\prime }}\left(a\right)=-{\int }_0^{2\pi }\frac{cosx}{{(acosx+bsinx)}^2}dx$$$$butf\left(a\right)=0\Rightarrow f^{{}^{\prime }}\left(a\right)=0\Rightarrow {\int }_0^{2\pi }\frac{cosx}{{(acosx+bsinx)}^2}dx=0alsowehave$$$${\int }_0^{2\pi }\frac{sinx}{{(acosx+bsinx)}^2}dx=0$$$$3){\int }_0^{2\pi }\frac{dx}{2cosx+\sqrt{3}sinx}=0.$$

Answered by tanmay last updated on 12/May/19

$$\int \frac{dx}{acosx+bsinx}$$$$\int \frac{dx}{rsin\alpha cosx+rcos\alpha sinx}$$$$\frac{1}{r}\int \frac{dx}{sin(\alpha +x)}$$$$\frac{1}{r}\int cosec(x+\alpha )dx$$$$\frac{1}{r}\times lntan\left(\frac{x+\alpha }{2}\right)+c$$$$\frac{1}{\sqrt{a^2+b^2}}lntan\left(\frac{x+{tan}^{-1}\left(\frac{a}{b}\right)}{2}\right)+c$$$$nowproceed$$

Answered by tanmay last updated on 13/May/19

$$2)\int \frac{cosx}{{(acosx+bsinx)}^2}$$$$cosx=A(acosx+bsinx)+B\times \frac{d}{dx}(acosx+bsinx)$$$$cosx=A(acosx+bsinx)+B\times (-asinx+bcosx)$$$$cosx=(Aa+bB)cosx+(Ab-Ba)sinx$$$$Ab-Ba=0$$$$\frac{A}{a}=\frac{B}{b}=k\to A=akandB=bk$$$$Aa+Bb=1$$$$a^{2}k+b^{2}k=1$$$$k=\frac{1}{a^2+b^2}soA=\frac{a}{a^2+b^2}andB=\frac{b}{a^2+b^2}$$$$now$$$$\int \frac{cosxdx}{{(acosx+bsinx)}^2}$$$$\int \left[\frac{A(acosx+bsinx)+B\times \frac{d}{dx}\frac{(acosx+bsinx)}{}}{{(acosx+bsinx)}^2}\right]dx$$$$\int \frac{A}{(acosx+bsinx)}dx+B\int \frac{d(acosx+bsinx)}{{(acosx+bsinx)}^2}$$$$=\frac{a}{a^2+b^2}\times \frac{1}{\sqrt{a^2+b^2}}lntan\left(\frac{x+{tan}^{-1}\left(\frac{a}{b}\right)}{2}\right)+\frac{b}{a^2+b^2}\times \frac{-1}{(acosx+bsinx)}+c$$$$nowplsputtheupperandlowerlimit$$$$for\int \frac{sinxdx}{{(acosx+bsinx)}^2}\to samemethod...$$$$$$$$$$$$$$$$$$$$$$

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