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  1. ∫ (csc√x)/√x dx put u = √x, then du = 1/(2√x) dx ==> 2 du = 1/√x dx. ... you remember the anti-derivative of the cosecant as a formula---it is -ln|csc(u) + cot(u)|. If not, there is a way to arrive at this. Multiply...

    1 Answers · Science & Mathematics · 19/11/2010

  2. csc(arccos(2 / sqrt(13)) Remember that: sin(t)^2 + cos(t)^2 = 1 and sin(t) = 1 / csc(t) So: (1/csc(t))^2 + cos(t)^2 = 1 (1/csc(arccos(2/sqrt(13...

    1 Answers · Science & Mathematics · 26/02/2012

  3. cscθ +cotθ = 1 1/sinθ + cosθ/sinθ = 1 1 + cosθ = sinθ sinθ - cosθ = 1 ...cosθ = 0 θ = {π/2, 3π/2, 5π/2....} Check: If sinθ=0, cscθ and cotθ are undefined. If θ=π/2 cscθ = 1 cotθ =0 cscθ+cot...

    1 Answers · Science & Mathematics · 06/05/2008

  4. csc^6x dx = change the sign: - ∫ (- csc^6x) dx = rewrite the integrand as: - ∫ (csc^4x) (- csc^2x) dx = rewrite (csc^4x) in terms of cotx: - ∫ (csc^2x)^2 (- csc...

    3 Answers · Science & Mathematics · 26/12/2008

  5. ∫ cot²x csc x dx = rewrite cot²x as (csc²x - 1): ∫ (csc²x - 1) csc... csc x - csc x) dx = and break it up into: ∫ csc²x csc x dx - ∫ csc x dx = as for the first integral, let: csc²x dx = dv → - cot x = v...

    3 Answers · Science & Mathematics · 29/06/2008

  6. csc^3(x) dx integrate by parts let u = csc x : du = - csc x cot x dv = csc^2(x) dx : v = - cot x...dx = - csc x cot x + ∫ csc x dx => 2 ∫csc^3(x) dx = - csc x cot x - ln I csc x + cot x I + c ∫csc...

    2 Answers · Science & Mathematics · 01/01/2010

  7. [csc θ / (csc θ - 1)] + [csc θ / (csc θ + 1)] = 2 sec²θ ~~~~~~~~~~~~~~~~~~~~~~~~~ METHOD... the LCD. (csc θ - 1) * (csc θ + 1) = (csc θ - 1)(csc θ + 1) Multiply each fraction on the LHS as ...

    4 Answers · Science & Mathematics · 04/11/2008

  8. f = √[tan(csc√x)/(sin^3(x) cot(2x))] df/dx = 1/2 * 1/√[tan(csc√x)/(sin^3x cot2x)] * d[tan(csc√x)/(sin^3x cot2x)]/dx d[tan(csc√x)/(sin...3sin^2x * cosx * cot2x - 2csc^2(x) * sin^3x d(csc√x)/dx = -1/(2√x) * cot(√x)csc(√x) d...

    1 Answers · Science & Mathematics · 15/12/2013

  9. cot x / (csc x + 1). Multiply by (csc x - 1) / (csc x -1). Basically that is the same as multiplying by 1. [cot x /(csc x+1)] x [(csc x -1)(csc x -1)] = [(cot x csc x - cot x) / cot^2 x] Then...

    3 Answers · Science & Mathematics · 21/02/2008

  10. Take the integral: ∫ (csc(2/x) + sec(2/x))²/x² dx Expanding the integrand (csc(2/x...(2/x))/x² dx + ∫ (sec²(2/x))/x² dx +2 ∫ (csc(2/x) sec(2/x))/x² dx For the integrand (csc(2/x) sec(2/x))/...

    3 Answers · Science & Mathematics · 24/12/2012

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