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Question Number 118891 by cantor last updated on 20/Oct/20

$${\int }_0^{\pi }\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(3^{\mathit{c}\mathit{o}\mathit{s}\mathit{x}}\right)\mathit{d}\mathit{x}=???$$$$$$$$\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}\mathit{h}\mathit{e}\mathit{l}\mathit{p}$$

Answered by Dwaipayan Shikari last updated on 20/Oct/20

$${\int }_0^{\pi }{tan}^{-1}\left(3^{cosx}\right)dx={\int }_0^{\pi }{tan}^{-1}\left(3^{-cosx}\right)dx=I(Using{\int }_0^{\pi }f\left(x\right)dx={\int }_0^{\pi }f(\pi -x)dx$$$$2I={\int }_0^{\pi }{tan}^{-1}\left(\frac{3^{cosx}+3^{-cosx}}{1-3^{cosx-cosx}}\right)dx$$$$2I={\int }_0^{\pi }{tan}^{-1}\left(\frac{3^{cosx}+3^{-cosx}}{z}\right)dxz\to 0{tan}^{-1}\left(\frac{3^{cosx}+3^{-cosx}}{z}\right)=\frac{\pi }{2}$$$$2I={\int }_0^{\pi }\frac{\pi }{2}dx$$$$I=\frac{{\pi }^2}{4}$$$$$$

Commented by mnjuly1970 last updated on 20/Oct/20

$$letmeexplain.$$$$note::{tan}^{-1}\left(x\right)+{tan}^{-1}\left(y\right)={tan}^{-1}\left(\frac{x+y}{1-xy}\right)$$$$I={\int }_0^{\pi }{tan}^{-1}\left(3^{cos\left(x\right)}\right)dx.....\left(1\right)$$$$I\stackrel{[{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx]}{=}{\int }_0^{\pi }{tan}^{-1}\left(3^{-cos\left(x\right)}\right)...\left(2\right)$$$$\therefore \left(1\right)+\left(2\right)::$$$$2I\stackrel{note}{=}{\int }_0^{\pi }{tan}^{-1}\left(\frac{3^{cos\left(x\right)}+3^{-cos\left(x\right)}}{[1-3^{cos\left(x\right)}.3^{-cos\left(x\right)}]=0}\right)$$$$$$$$={\int }_0^{\pi }[{tan}^{-1}\left(\mathrm{\infty }\right)=\frac{\pi }{2}]dx=\frac{{\pi }^2}{2}$$$$\therefore I=\frac{{\pi }^2}{4}✓✓$$$$...m.n.1970...$$

Commented by cantor last updated on 20/Oct/20

$$\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}\mathit{i}{\mathit{d}\mathit{o}\mathit{n}}^{\prime }\mathit{t}\mathit{u}\mathit{n}\mathit{d}\mathit{e}\mathit{r}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{d}$$$$\mathit{t}\mathit{h}\mathit{e}{1}^{\mathit{s}\mathit{t}}\mathit{l}\mathit{i}\mathit{n}\mathit{e}\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}\mathit{e}\mathit{x}\mathit{p}\mathit{l}\mathit{a}\mathit{i}\mathit{n}\mathit{t}$$

Commented by benjo_mathlover last updated on 20/Oct/20

$$replacexby\pi -x$$

Answered by mindispower last updated on 20/Oct/20

$$={\int }_0^{\pi }arctan\left(3^{cos(\pi -x)}\right)dx=\int arctan\left(\frac{1}{3^{cos\left(x\right)}}\right)$$$$arctan\left(\frac{1}{x}\right)=\frac{\pi }{2}-arctan\left(x\right),x>0$$$$.....$$

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