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 93098 by Rio Michael last updated on 10/May/20

$$\mathrm{find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{to}$$$$\int \frac{1}{a\sin x+b\cos x}dx\mathrm{and}\int \frac{1}{a\cos x-b\sin x}dx$$$$\mathrm{where}a,b\mathrm{are}\mathrm{constants}.$$$$$$

Commented by prakash jain last updated on 10/May/20

$$a\sin x+b\cos x$$$$=\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\sin x+\frac{b}{\sqrt{a^2+b^2}}\cos x)$$$$\frac{a}{\sqrt{a^2+b^2}}=\cos \alpha$$$$\frac{b}{\sqrt{a^2+b^2}}=\sin \alpha$$$$\sqrt{a^2+b^2}=\sqrt{a^2+b^2}\sin (\alpha +x)$$$$\frac{1}{a\sin x+b\cos x}=\sqrt{a^2+b^2}\mathrm{cosec}(x+\alpha )$$$$\mathrm{now}\mathrm{use}\mathrm{cosec}\mathrm{formula}\mathrm{for}\mathrm{integration}.$$$$------$$$$\mathrm{Same}\mathrm{procedure}\mathrm{can}\mathrm{be}\mathrm{use}$$$$\frac{1}{a\sin x-b\cos x}$$

Commented by Rio Michael last updated on 10/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

Commented by abdomathmax last updated on 11/May/20

$$I=\int \frac{dx}{acosx+bsinx}wesupposea\ne 0\Rightarrow$$$$I=\frac{1}{a}\int \frac{dx}{cosx+\lambda sinx}=A_{\lambda }(\lambda =\frac{b}{a})$$$$changementtan\left(\frac{x}{2}\right)=tgive$$$$A_{\lambda }=\frac{1}{a}\int \frac{2dt}{(1+t^2)(\frac{1-t^2}{1+t^2}+\lambda \frac{2t}{1+t^2})}$$$${aA}_{\lambda }=\int \frac{2dt}{1-t^2+2\lambda t}=-\int \frac{2dt}{t^2-2\lambda t-1}$$$$letsolve{t}^2-2\lambda t-1=0$$$${\mathrm{\Delta }}^{{}^{\prime }}={\lambda }^2+1\Rightarrow t_1=\lambda +\sqrt{1+{\lambda }^2}$$$$t_2=\lambda -\sqrt{1+{\lambda }^2}$$$${aA}_{\lambda }=-\int \frac{2dt}{(t-t_1)(t-t_2)}$$$$=-\frac{2}{2\sqrt{1+{\lambda }^2}}\int (\frac{1}{t-t_1}-\frac{1}{t-t_2})dt$$$$=\frac{1}{\sqrt{1+{\lambda }^2}}ln\mid \frac{t-t_2}{t-t_1}\mid +C$$$$=\frac{1}{\sqrt{1+{\lambda }^2}}ln\mid \frac{t-\lambda +\sqrt{1+{\lambda }^2}}{t-\lambda -\sqrt{1+{\lambda }^2}}\mid +C$$$$=\frac{1}{\sqrt{1+\frac{b^2}{a^2}}}ln\mid \frac{t-\frac{b}{a}+\sqrt{1+\frac{b^2}{a^2}}}{t-\frac{b}{a}-\sqrt{1+\frac{b^2}{a^2}}}\mid +C$$$$=\frac{\mid a\mid }{\sqrt{a^2+b^2}}ln\mid \frac{\frac{at-b}{a}+\frac{\sqrt{a^2+b^2}}{\mid a\mid }}{\frac{at-b}{a}-\frac{\sqrt{a^2+b^2}}{\mid a\mid }}\mid +C$$$$ifwetakea>0weget$$$$I=\frac{1}{\sqrt{a^2+b^2}}ln\mid \frac{atan\left(\frac{x}{2}\right)-b+\sqrt{a^2+b^2}}{atan\left(\frac{x}{2}\right)-b-\sqrt{a^2+b^2}}\mid +C$$$$=$$$$=$$

Commented by Rio Michael last updated on 11/May/20

$$\mathrm{wow}\mathrm{sir}\mathrm{spendid}$$

Commented by mathmax by abdo last updated on 11/May/20

$$thankx$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com