3.47 \(\int \sqrt{1-\cos ^2(x)} \, dx\)

Optimal. Leaf size=12 \[ \sqrt{\sin ^2(x)} (-\cot (x)) \]

[Out]

-(Cot[x]*Sqrt[Sin[x]^2])

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Rubi [A]  time = 0.0158364, antiderivative size = 12, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3176, 3207, 2638} \[ \sqrt{\sin ^2(x)} (-\cot (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - Cos[x]^2],x]

[Out]

-(Cot[x]*Sqrt[Sin[x]^2])

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \sqrt{1-\cos ^2(x)} \, dx &=\int \sqrt{\sin ^2(x)} \, dx\\ &=\left (\csc (x) \sqrt{\sin ^2(x)}\right ) \int \sin (x) \, dx\\ &=-\cot (x) \sqrt{\sin ^2(x)}\\ \end{align*}

Mathematica [A]  time = 0.0061805, size = 12, normalized size = 1. \[ \sqrt{\sin ^2(x)} (-\cot (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - Cos[x]^2],x]

[Out]

-(Cot[x]*Sqrt[Sin[x]^2])

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Maple [A]  time = 0.463, size = 13, normalized size = 1.1 \begin{align*} -{\cos \left ( x \right ) \sin \left ( x \right ){\frac{1}{\sqrt{ \left ( \sin \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-cos(x)^2)^(1/2),x)

[Out]

-sin(x)*cos(x)/(sin(x)^2)^(1/2)

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Maxima [A]  time = 1.44539, size = 14, normalized size = 1.17 \begin{align*} -\frac{1}{\sqrt{\tan \left (x\right )^{2} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(x)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/sqrt(tan(x)^2 + 1)

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Fricas [A]  time = 1.89958, size = 12, normalized size = 1. \begin{align*} -\cos \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(x)^2)^(1/2),x, algorithm="fricas")

[Out]

-cos(x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{1 - \cos ^{2}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(x)**2)**(1/2),x)

[Out]

Integral(sqrt(1 - cos(x)**2), x)

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Giac [B]  time = 1.271, size = 32, normalized size = 2.67 \begin{align*} -\frac{2 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, x\right )^{3} + \tan \left (\frac{1}{2} \, x\right )\right )}{\tan \left (\frac{1}{2} \, x\right )^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(x)^2)^(1/2),x, algorithm="giac")

[Out]

-2*sgn(tan(1/2*x)^3 + tan(1/2*x))/(tan(1/2*x)^2 + 1)