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Question Number 30775 by abdo imad last updated on 25/Feb/18

$$let\alpha \in ]0,\pi [calculate{\int }_0^{\frac{\pi }{2}}\frac{dx}{2(cos\alpha +chx)}.$$

Commented by prof Abdo imad last updated on 25/Feb/18

$$wehaveI={\int }_0^{\frac{\pi }{2}}\frac{dx}{2(cos\alpha +\frac{e^x+e^{-x}}{2})}$$$$={\int }_0^{\frac{\pi }{2}}\frac{dx}{2cos\alpha +e^x+e^{-x}}thech.e^x=t$$$$I={\int }_1^{e^{\frac{\pi }{2}}}\frac{1}{2cos\alpha +t+\frac{1}{t}}\frac{dt}{t}$$$$={\int }_1^{e^{\frac{\pi }{2}}}\frac{dt}{2tcos\alpha +t^2+1}={\int }_1^{e^{\frac{\pi }{2}}}\frac{dt}{t^2+2tcos\alpha +1}$$$$={\int }_0^{e^{\frac{\pi }{2}}}\frac{dt}{{(t+cos\alpha )}^2+{sin}^2\alpha }letusethech.$$$$t+cos\alpha =usin\alpha \Rightarrow u=\frac{1}{sin\alpha }(t+cos\alpha )$$$$I={\int }_{cotan\left(\alpha \right)}^{\frac{e^{\frac{\pi }{2}}+cos\alpha }{sin\alpha }}\frac{sin\alpha du}{{sin}^2\alpha (1+u^2)}$$$$=\frac{1}{sin\alpha }{\left[arctanu\right]}_{cotan\alpha }^{\frac{e^{\frac{\pi }{2}}+cos\alpha }{sin\alpha }}$$$$=\frac{1}{sin\alpha }(arctan\left(\frac{e^{\frac{\pi }{2}}+cos\alpha }{sin\alpha }\right)-arctan\left(cotan\alpha \right)).$$$$$$

Answered by sma3l2996 last updated on 25/Feb/18

$$A={\int }_0^{\pi /2}\frac{dx}{2(cos\alpha +chx)}$$$$lett=th\left(x/2\right)\Rightarrow dt=\frac{1}{2}(1-{th}^2\left(x/2\right))dx$$$$dx=\frac{2}{1-t^2}dt$$$$ch\left(x\right)=2{ch}^2\left(x/2\right)-1=\frac{2}{1-{th}^2\left(x/2\right)}-1=\frac{2}{1-t^2}-1=\frac{1+t^2}{1-t^2}$$$$A={\int }_0^{th\left(\pi /4\right)}\frac{1}{2(cos\alpha +\frac{1+t^2}{1-t^2})}\times \left(\frac{2dt}{1-t^2}\right)$$$$={\int }_0^{th\left(\pi /4\right)}\frac{dt}{(1-t^2)cos\alpha +1+t^2}={\int }_0^{th\left(\pi /4\right)}\frac{dt}{t^2(1-cos\alpha )+1+cos\alpha }$$$$A={\int }_0^{th\left(\pi /4\right)}\frac{dt}{(1+cos\alpha )(\left(\frac{1-cos\alpha }{1+cos\alpha }\right)t^2+1)}$$$${let}^{\prime }sputu={\left(\frac{1-cos\alpha }{1+cos\alpha }\right)}^{1/2}t\Rightarrow dt=\left(\frac{1+cos\alpha }{1-cos\alpha }\right)du$$$$A=\frac{1}{{(1-cos\alpha )}^{1/2}{(1+cos\alpha )}^{1/2}}{\int }_0^a\frac{du}{u^2+1}$$$$A=\frac{1}{{(1-{cos}^2\alpha )}^{1/2}}\times {\left[arctan\left(\frac{1-cos\alpha }{sin\alpha }t\right)\right]}_0^{th\left(\pi /4\right)}$$$$A=\frac{1}{sin\alpha }arctan\left(\frac{1-cos\alpha }{sin\alpha }th\left(\pi /4\right)\right)$$$$$$

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