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Question Number 71717 by peter frank last updated on 19/Oct/19

Commented by mathmax by abdo last updated on 19/Oct/19

$$letI=\int \frac{dx}{cos(x-a)cos(x-b)}\Rightarrow I=\int \frac{2dx}{cos(2x-a-b)+cos(a-b)}$$$${=}_{2x-(a+b)=t}\int \frac{dt}{cost+\lambda }with\lambda =cos(a-b)changement$$$$tan\left(\frac{t}{2}\right)=ugive\int \frac{dt}{cost+\lambda }=\int \frac{1}{\frac{1-u^2}{1+u^2}+\lambda }\frac{2du}{1+u^2}$$$$=\int \frac{2du}{1-u^2+\lambda (1+u^2)}=\int \frac{2du}{(\lambda -1)u^2+1+\lambda }=\int \frac{-2du}{(1-\lambda )u^2-(1+\lambda )}$$$$=\frac{-2}{1-\lambda }\int \frac{du}{u^2-\frac{1+\lambda }{1-\lambda }}(lookthat\frac{1+\lambda }{1-\lambda }>0)$$$${=}_{u=\sqrt{\frac{1+\lambda }{1-\lambda }}z}\frac{-2}{1-\lambda }\frac{1-\lambda }{1+\lambda }\int \frac{1}{z^2-1}\frac{\sqrt{1+\lambda }}{\sqrt{1-\lambda }}dz$$$$=\frac{-1}{\sqrt{1-{\lambda }^2}}\int (\frac{1}{z-1}-\frac{1}{z+1})dz=\frac{-1}{\sqrt{1-{\lambda }^2}}ln\mid \frac{z-1}{z+1}\mid +C$$$$=\frac{-1}{\sqrt{1-{cos}^2(a-b)}}\times ln\mid \frac{\sqrt{\frac{1-\lambda }{1+\lambda }}tan\left(\frac{t}{2}\right)-1}{\sqrt{\frac{1-\lambda }{1+\lambda }}tan\left(\frac{t}{2}\right)+1}\mid +C$$$$=\frac{-1}{\mid sin(a-b)\mid }\times ln\mid \frac{tan\left(\frac{a-b}{2}\right)tan(x-\frac{a+b}{2})-1}{tan\left(\frac{a-b}{2}\right)tan(x-\frac{a+b}{2})+1}\mid +Cwitha-b\ne k\pi$$$$k\in Z$$$$$$

Commented by peter frank last updated on 20/Oct/19

$$thankyou$$$$$$

Commented by mathmax by abdo last updated on 23/Oct/19

$$youarewelcome.$$

Answered by Tanmay chaudhury last updated on 19/Oct/19

$$\frac{1}{sin(a-b)}\int \frac{sin\{(x-b)-(x-a)\}}{cos(x-a)cos(x-b)}dx$$$$\frac{1}{sin(a-b)}\int \frac{sin(x-b)cos(x-a)-cos(x-b)sin(x-a)}{cos(x-a)cos(x-b)}dx$$$$\frac{1}{sin(a-b)}\int [tan(x-b)-tan(x-a)]dx$$$$\frac{1}{sin(a-b)}[lnsec(x-b)-lnsec(x-a)]+c$$$$\frac{1}{sin(a-b)}ln\left(\frac{sec(x-b)}{sec(x-a)}\right)+c$$

Commented by $@ty@m123 last updated on 19/Oct/19

$$Quiteeasierbuttricky.$$

Commented by peter frank last updated on 19/Oct/19

$$thankyouverymuch$$$$$$

Commented by Prithwish sen last updated on 19/Oct/19

$$\mathrm{tanmoy}\mathrm{sir}\mathrm{pujo}\mathrm{kamon}\mathrm{katlo}.\mathrm{subho}\mathrm{bijaya}.$$$$\mathrm{kothae}\mathrm{chilen}?\mathrm{bhalo}\mathrm{achen}?$$

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