Given-that-y-sin-x-1-cos-x-find-dy-dx-Evaluate-1-2-x-4-dx-

Question Number 33005 by Rio Mike last updated on 09/Apr/18

G i v e n t h a t

y = \frac{s i n x}{1 + c o s x} f i n d \frac{d y}{d x}

E v a l u a t e

\int_{1}^{2} (x + 4) d x

Commented by abdo imad last updated on 09/Apr/18

\frac{d y}{d x} = \frac{c o s x (1 + c o s x) + {s i n}^{2} x}{{(1 + c o s x)}^{2}} = \frac{c o s x + {s i n}^{2} x + {c o s}^{2} x}{{(1 + c o s x)}^{2}} .

= \frac{1 + c o s x}{{(1 + c o s x)}^{2}} = \frac{1}{1 + c o s x} f o r x \in D_{y} .

★ \int_{1}^{2} (x + 4) d x = {[\frac{x^{2}}{2} + 4 x]}_{1}^{2} = 2 + 8 - \frac{1}{2} - 4

= \frac{3}{2} + 4 = \frac{11}{2} .