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Question Number 55615 by Abdo msup. last updated on 28/Feb/19

$$letF\left(\alpha \right)={\int }_{\alpha }^{1+{\alpha }^2}\frac{sin\left(\alpha x\right)}{1+\alpha x^2}dx$$$$1)calculate\frac{dF}{d\alpha }\left(\alpha \right)$$$$2)calculate{lim}_{\alpha \to 0}F\left(\alpha \right)$$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Feb/19

$$\frac{dF}{d\alpha }={\int }_{\alpha }^{1+{\alpha }^2}\frac{\partial }{\partial \alpha }\left(\frac{sin\alpha x}{1+\alpha x^2}\right)dx+\frac{sin\alpha (1+{\alpha }^2)}{1+\alpha {(1+\alpha )}^2}\frac{d}{d\alpha }(1+{\alpha }^2)-\frac{sin\alpha \left(\alpha \right)}{1+\alpha {\left(\alpha \right)}^2}\times \frac{d}{d\alpha }\left(\alpha \right)$$$$={\int }_{\alpha }^{1+{\alpha }^2}\frac{(1+\alpha x^2)\times cos\alpha x\times x-sin\alpha x(0+x^2)}{{(1+\alpha x^2)}^2}dx+\frac{sin(\alpha +{\alpha }^3)}{1+\alpha +2{\alpha }^2+{\alpha }^3}\times \left(2\alpha \right)-\frac{sin{\alpha }^2}{1+{\alpha }^3}\times 1$$$$={\int }_{\alpha }^{1+{\alpha }^2}\frac{xcos\alpha x+\alpha x^{3}cos\alpha x^2-x^{2}sin\alpha x}{{(1+\alpha x^2)}^2}dx+\frac{2\alpha sin(\alpha +{\alpha }^3)}{1+\alpha +2{\alpha }^2+{\alpha }^3}-\frac{sin{\alpha }^2}{1+{\alpha }^3}$$$$wait...$$

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