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integration of (sinx)^1/2 is not possible.so integration of root sinx is impossible

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โˆ™ 2009-09-21 14:25:19
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: Integration of root sinx
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Integration of sinx is?

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Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?

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What is answer of integral of square root of sinx?

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How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?

(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx


How do you differentiate sine squared x?

Using the Chain Rule :derivative of (sinx)2 = 2(sinx)1 * (derivative of sinx)d/dx (Sinx)2 = 2(sinx)1 * [d/dx (Sinx)]d/dx (Sinx)2 = 2(sinx) * (cosx)d/dx (Sinx)2 = 2 (sinx) * (cosx)d/dx (Sinx)2 = 2 sin(x) * cos(x)


How do you verify the identity sinx cscx 1?

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It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)


Can you Show 1 over sinx cosx - cosx over sinx equals tanx?

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What is 1- sinx?

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