g(x)=cos(2arcsinx) calculate (dg/dx) and (d^2 g/dx^2 ) 2)find ∫


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 146901 by mathmax by abdo last updated on 16/Jul/21

$g (x) = \cos (2 arcsinx)$ $calculate \frac{dg}{dx} and \frac{d^{2} g}{{dx}^{2}}$ $2) find \int_{- \frac{1}{2}}^{\frac{1}{2}} g (x) dx$
	Answered by ArielVyny last updated on 18/Jul/21

	$\frac{d g}{d x} = - 2 \frac{1}{\sqrt{1 - x^{2}}} s i n (2 a r c s i n x)$ $\frac{d^{2} g}{{d x}^{2}} = - 2 [- \frac{\frac{- 2 x}{2 \sqrt{1 - x^{2}}}}{1 - x^{2}} s i n (2 a r c s i n x) - 2 \frac{1}{\sqrt{1 - x^{2}}} c o s (2 a r s i n x)]$ $= - 2 [- \frac{- x}{\sqrt{1 - x^{2}}} \times \frac{1}{1 - x^{2}} s i n (2 a r s i n x) - 2 \frac{1}{\sqrt{1 - x^{2}}} c o s (2 a r c s i n x)]$ $\frac{d^{2} g}{{d x}^{2}} = - 2 [\frac{x s i n (2 a r c s i n x)}{{(1 - x^{2})}^{\frac{3}{2}}} - 2 \frac{c o s (2 a r c s i n x)}{\sqrt{1 - x^{2}}}]$ $2 . . . \int_{- \frac{1}{2}}^{\frac{1}{2}} c o s [2 a r c s i n x] d x$ $f (x) = c o s (2 a r c s i n x) e s t p a i r e d o n c$ $\int_{- \frac{1}{2}}^{\frac{1}{2}} c o s (2 a r c s i n x) d x = 2 \int_{0}^{\frac{1}{2}} c o s (2 a r c s i n x) d x$ $o n p o s e d u = 1 \to u = x$ $v = c o s (2 a r c s i n x) \to d v = - 2 \frac{1}{\sqrt{1 - x^{2}}} s i n (2 a r c s i n x)$ $I = 2 [x c o s (2 a r c s i n x)] + 2 \int_{0}^{\frac{1}{2}} \frac{2 x}{\sqrt{1 - x^{2}}} s i n (2 a r c s i n x) d x$ $d u = \frac{2 x}{\sqrt{1 - x^{2}}} \to u = - \sqrt{1 - x^{2}}$ $v = s i n (2 a r c s i n x) \to d v = 2 \frac{1}{\sqrt{1 - x^{2}}} c o s (2 a r s i n x)$ $I = {[2 x c o s (2 a r c s i n x)]}_{0}^{1} - {[2 \sqrt{1 - x^{2}} s i n (2 a r c s i n x)]}_{0}^{1} + 2 \int_{0}^{1} \frac{2}{\sqrt{1 - x^{2}}} c o s (2 a r c s i n x)$ $I = {[2 x c o s (2 a r c s i n x) - 2 \sqrt{1 - x^{2}} s i n (2 a r c s i n x) + 2 s i n (2 a r c s i n x)]}_{0}^{1}$ $I = - 2$ $M r m a t h m a x s t i l l c h e k m y w o r k p l e a s e$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum