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Question Number 32712 by caravan msup abdo. last updated on 31/Mar/18

$$calculate{\int }_0^{\frac{\pi }{2}}\frac{dt}{1+a{cos}^{2}t}.$$

Commented by abdo imad last updated on 03/Apr/18

$$letputF\left(a\right)={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+a{cos}^{2}t}$$$$F\left(a\right)={\int }_0^{\frac{\pi }{2}}\frac{dt}{1+a\frac{1+cos\left(2t\right)}{2}}={\int }_0^{\frac{\pi }{2}}\frac{2dt}{2+a+acos\left(2t\right)}$$$${=}_{2t=u}{\int }_0^{\pi }\frac{2}{2+a+acos\left(u\right)}\frac{du}{2}$$$$={\int }_0^{\pi }\frac{du}{2+a+acosu}butthech.tan\left(\frac{u}{2}\right)=xgive$$$$F\left(a\right)={\int }_0^{+\mathrm{\infty }}\frac{1}{2+a+a\frac{1-x^2}{1+x^2}}\frac{2dx}{1+x^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{2dx}{(2+a)(1+x^2)+a(1-x^2)}$$$$={\int }_0^{\mathrm{\infty }}\frac{2dx}{2+a+(2+a)x^2+a-{ax}^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{2dx}{2+2a+2x^2}={\int }_0^{\mathrm{\infty }}\frac{dx}{1+a+x^2}$$$$case11+a>0\Rightarrow F\left(a\right){=}_{x=\sqrt{1+a}u}{\int }_0^{\mathrm{\infty }}\frac{\sqrt{1+a}du}{(1+a)(1+u^2)}$$$$=\frac{\pi }{2\sqrt{1+a}}$$$$case21+a<0\Rightarrow F\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{x^2-{\left(\sqrt{-(1+a)}\right)}^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{dx}{(x-\alpha )(x+\alpha )}(\alpha =\sqrt{-1-a})$$$$=\frac{1}{2\alpha }{\int }_0^{\mathrm{\infty }}(\frac{1}{x-\alpha }-\frac{1}{x+\alpha })dx=\frac{1}{2\alpha }{[ln\mid \frac{x-\alpha }{x+\alpha }\mid ]}_0^{+\mathrm{\infty }}=0$$

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