Інтеграл від 1x2\sqrt{1 - x^{2}}

Калькулятор знайде інтеграл/антипохідну від 1x2\sqrt{1 - x^{2}}, з показаними кроками.

Пов'язаний калькулятор: Інтегральний калькулятор

Розв'язок

Let x=sin(u)x=\sin{\left(u \right)}.

Then dx=(sin(u))du=cos(u)dudx=\left(\sin{\left(u \right)}\right)^{\prime }du = \cos{\left(u \right)} du (steps can be seen »).

Also, it follows that u=asin(x)u=\operatorname{asin}{\left(x \right)}.

Thus,

1x2=1sin2(u)\sqrt{1 - x^{2}} = \sqrt{1 - \sin^{2}{\left( u \right)}}

Use the identity 1sin2(u)=cos2(u)1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}:

1sin2(u)=cos2(u)\sqrt{1 - \sin^{2}{\left( u \right)}}=\sqrt{\cos^{2}{\left( u \right)}}

Assuming that cos(u)0\cos{\left( u \right)} \ge 0, we obtain the following:

cos2(u)=cos(u)\sqrt{\cos^{2}{\left( u \right)}} = \cos{\left( u \right)}

Тому,

1x2dx=cos2(u)du{\color{red}{\int{\sqrt{1 - x^{2}} d x}}} = {\color{red}{\int{\cos^{2}{\left(u \right)} d u}}}

Rewrite the cosine using the power reducing formula cos2(α)=cos(2α)2+12\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2} with α=u\alpha= u :

cos2(u)du=(cos(2u)2+12)du{\color{red}{\int{\cos^{2}{\left(u \right)} d u}}} = {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}

Apply the constant multiple rule cf(u)du=cf(u)du\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du with c=12c=\frac{1}{2} and f(u)=cos(2u)+1f{\left(u \right)} = \cos{\left(2 u \right)} + 1:

(cos(2u)2+12)du=((cos(2u)+1)du2){\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}} = {\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}

Integrate term by term:

(cos(2u)+1)du2=(1du+cos(2u)du)2\frac{{\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}}}{2} = \frac{{\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}}{2}

Apply the constant rule cdu=cu\int c\, du = c u with c=1c=1:

cos(2u)du2+1du2=cos(2u)du2+u2\frac{\int{\cos{\left(2 u \right)} d u}}{2} + \frac{{\color{red}{\int{1 d u}}}}{2} = \frac{\int{\cos{\left(2 u \right)} d u}}{2} + \frac{{\color{red}{u}}}{2}

Let v=2uv=2 u.

Then dv=(2u)du=2dudv=\left(2 u\right)^{\prime }du = 2 du (steps can be seen »), and we have that du=dv2du = \frac{dv}{2}.

So,

u2+cos(2u)du2=u2+cos(v)2dv2\frac{u}{2} + \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{2} = \frac{u}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{2}

Apply the constant multiple rule cf(v)dv=cf(v)dv\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv with c=12c=\frac{1}{2} and f(v)=cos(v)f{\left(v \right)} = \cos{\left(v \right)}:

u2+cos(v)2dv2=u2+(cos(v)dv2)2\frac{u}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{2} = \frac{u}{2} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{2}

The integral of the cosine is cos(v)dv=sin(v)\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}:

u2+cos(v)dv4=u2+sin(v)4\frac{u}{2} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{4} = \frac{u}{2} + \frac{{\color{red}{\sin{\left(v \right)}}}}{4}

Recall that v=2uv=2 u:

u2+sin(v)4=u2+sin((2u))4\frac{u}{2} + \frac{\sin{\left({\color{red}{v}} \right)}}{4} = \frac{u}{2} + \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{4}

Recall that u=asin(x)u=\operatorname{asin}{\left(x \right)}:

sin(2u)4+u2=sin(2asin(x))4+asin(x)2\frac{\sin{\left(2 {\color{red}{u}} \right)}}{4} + \frac{{\color{red}{u}}}{2} = \frac{\sin{\left(2 {\color{red}{\operatorname{asin}{\left(x \right)}}} \right)}}{4} + \frac{{\color{red}{\operatorname{asin}{\left(x \right)}}}}{2}

Тому,

1x2dx=sin(2asin(x))4+asin(x)2\int{\sqrt{1 - x^{2}} d x} = \frac{\sin{\left(2 \operatorname{asin}{\left(x \right)} \right)}}{4} + \frac{\operatorname{asin}{\left(x \right)}}{2}

Using the formulas sin(2asin(α))=2α1α2\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}, sin(2acos(α))=2α1α2\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}, cos(2asin(α))=12α2\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}, cos(2acos(α))=2α21\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1, sinh(2asinh(α))=2αα2+1\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}, sinh(2acosh(α))=2αα1α+1\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}, cosh(2asinh(α))=2α2+1\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1, cosh(2acosh(α))=2α21\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1, simplify the expression:

1x2dx=x1x22+asin(x)2\int{\sqrt{1 - x^{2}} d x} = \frac{x \sqrt{1 - x^{2}}}{2} + \frac{\operatorname{asin}{\left(x \right)}}{2}

Додайте константу інтегрування:

1x2dx=x1x22+asin(x)2+C\int{\sqrt{1 - x^{2}} d x} = \frac{x \sqrt{1 - x^{2}}}{2} + \frac{\operatorname{asin}{\left(x \right)}}{2}+C

Answer: 1x2dx=x1x22+asin(x)2+C\int{\sqrt{1 - x^{2}} d x}=\frac{x \sqrt{1 - x^{2}}}{2} + \frac{\operatorname{asin}{\left(x \right)}}{2}+C