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Question Number 147746 by 0731619 last updated on 23/Jul/21

Answered by mindispower last updated on 23/Jul/21

$$\frac{{tan}^{-}\left(x^2\right)}{{tan}^{-}\left(x\right)}<1$$$$x⩽x^2<1,andusing{tan}^{-}increasingfunction$$$$\mathrm{\forall }x\in ]a,b[x<1\Rightarrow xsin\left(x\right)<sin\left(x\right)$$$$\Rightarrow \frac{sin\left(x\right)}{x}>sin\left(x\right)$$$$\Rightarrow \frac{sin\left(x\right)}{x}.\frac{ta\stackrel{-}{n}\left(x^2\right)}{{tan}^{-}\left(\mathit{x}\right)}\mathit{d}\mathit{x}>\mathit{s}\mathit{i}\mathit{n}\left(\mathit{x}\right)$$$${\int }_a^b\frac{sin\left(x\right){tan}^{-}\left(x^2\right)}{x.{tan}^{-}\left(x\right)}dx>{\int }_a^{b}sin\left(x\right)dx=cos\left(a\right)-cos\left(b\right)$$

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