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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 \dots \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