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

l e t F (α) = \int_{α}^{1 + α^{2}} \frac{s i n (α x)}{1 + α x^{2}} d x

1) c a l c u l a t e \frac{d F}{d α} (α)

2) c a l c u l a t e {l i m}_{α \to 0} F (α)

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

\frac{d F}{d α} = \int_{α}^{1 + α^{2}} \frac{\partial}{\partial α} (\frac{s i n α x}{1 + α x^{2}}) d x + \frac{s i n α (1 + α^{2})}{1 + α {(1 + α)}^{2}} \frac{d}{d α} (1 + α^{2}) - \frac{s i n α (α)}{1 + α {(α)}^{2}} \times \frac{d}{d α} (α)

= \int_{α}^{1 + α^{2}} \frac{(1 + α x^{2}) \times c o s α x \times x - s i n α x (0 + x^{2})}{{(1 + α x^{2})}^{2}} d x + \frac{s i n (α + α^{3})}{1 + α + 2 α^{2} + α^{3}} \times (2 α) - \frac{s i n α^{2}}{1 + α^{3}} \times 1

= \int_{α}^{1 + α^{2}} \frac{x c o s α x + α x^{3} c o s α x^{2} - x^{2} s i n α x}{{(1 + α x^{2})}^{2}} d x + \frac{2 α s i n (α + α^{3})}{1 + α + 2 α^{2} + α^{3}} - \frac{s i n α^{2}}{1 + α^{3}}

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