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Question Number 167510 by Mathspace last updated on 18/Mar/22

$$explicitef\left(a\right)={\int }_0^{\frac{\pi }{2}}ln(a+{tan}^{2}x)dx$$$$a⩾2$$

Answered by ArielVyny last updated on 20/Mar/22

$$f\left(a\right)={\int }_0^{\frac{\pi }{2}}ln(a+{tan}^{2}x)dx$$$$f^{\prime }\left(a\right)={\int }_0^{\frac{\pi }{2}}\frac{1}{a+{tan}^{2}x}dx$$$$t=tanx\to dt=(1+t^2)dx$$$${\int }_0^1\frac{1}{a+t^2}.\frac{1}{1+t^2}dt={\int }_0^1\frac{1}{a+t^2}.[1-\frac{t^2}{1+t^2}]dt$$$$\frac{1}{a}{\int }_0^1\frac{1}{1+{\left(\frac{t}{\sqrt{a}}\right)}^2}dt-{\int }_0^1\frac{t^2}{(1+t^2)(a+t^2)}$$$$\frac{\sqrt{a}}{a}\left[arctg\left(\frac{1}{\sqrt{a}}\right)\right]-{\int }_0^1\frac{t^2}{(1+t^2)(a+t^2)}dt$$$$\frac{t^2}{(1+t^2)(a+t^2)}=\frac{\alpha }{1+t^2}+\frac{\beta }{a+t^2}=\frac{\alpha (a+t^2)+\beta (1+t^2)}{(1+t^2)(a+t^2)}$$$$=\frac{t^2(\alpha +\beta )+\alpha a+\beta }{(1+t^2)(a+t^2)}$$$$\alpha +\beta =1et\alpha a+\beta =0$$$$\alpha -\alpha a=1\to \alpha (1-a)=1\to \alpha =-\frac{1}{a-1}$$$$\beta =\frac{a}{a-1}$$$$f^{\prime }\left(a\right)={\int }_0^1\frac{1}{a+t^2}-[-\frac{1}{a-1}{\int }_0^1\frac{1}{1+t^2}+\frac{a}{a-1}{\int }_0^1\frac{1}{a+t^2}dt]$$$$=(1+\frac{a}{a-1}){\int }_0^1\frac{1}{a+t^2}+\frac{1}{a-1}.\frac{\pi }{4}$$$$=\frac{\pi }{4(a-1)}+(\frac{1}{a}+\frac{1}{a-1}){\int }_0^1\frac{1}{1+\frac{t^2}{a}}$$$${\int }_0^1\frac{1}{1+\frac{t^2}{a}}=\sum _{n⩾0}{\int }_0^1{(-1)}^n{\left(\frac{t^2}{a}\right)}^n=\sum _{n⩾0}\frac{{(-1)}^n}{a^n}$$$$f^{\prime }\left(a\right)=\frac{\pi }{4(a-1)}+(\frac{1}{a}+\frac{1}{a-1})\sum _{n⩾0}\frac{{(-1)}^n}{a^n}$$$$f\left(a\right)=\frac{\pi }{4}ln(a-1)+\sum _{n⩾1}{(-1)}^{n+1}\frac{1}{{na}^n}+\sum _{n⩾0}\int \frac{{(-1)}^n}{a^{n+1}-a^n}da$$$$$$

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