Розв'язок Let x = sin ( u ) x=\sin{\left(u \right)} x = sin ( u ) .
Then d x = ( sin ( u ) ) ′ d u = cos ( u ) d u dx=\left(\sin{\left(u \right)}\right)^{\prime }du = \cos{\left(u \right)} du d x = ( sin ( u ) ) ′ d u = cos ( u ) d u (steps can be seen » ).
Also, it follows that u = asin ( x ) u=\operatorname{asin}{\left(x \right)} u = asin ( x ) .
Thus,
1 − x 2 = 1 − sin 2 ( u ) \sqrt{1 - x^{2}} = \sqrt{1 - \sin^{2}{\left( u \right)}} 1 − x 2 = 1 − sin 2 ( u )
Use the identity 1 − sin 2 ( u ) = cos 2 ( u ) 1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)} 1 − sin 2 ( u ) = cos 2 ( u ) :
1 − sin 2 ( u ) = cos 2 ( u ) \sqrt{1 - \sin^{2}{\left( u \right)}}=\sqrt{\cos^{2}{\left( u \right)}} 1 − sin 2 ( u ) = cos 2 ( u )
Assuming that cos ( u ) ≥ 0 \cos{\left( u \right)} \ge 0 cos ( u ) ≥ 0 , we obtain the following:
cos 2 ( u ) = cos ( u ) \sqrt{\cos^{2}{\left( u \right)}} = \cos{\left( u \right)} cos 2 ( u ) = cos ( u )
Тому,
∫ 1 − x 2 d x = ∫ cos 2 ( u ) d u {\color{red}{\int{\sqrt{1 - x^{2}} d x}}} = {\color{red}{\int{\cos^{2}{\left(u \right)} d u}}} ∫ 1 − x 2 d x = ∫ c o s 2 ( u ) d u
Rewrite the cosine using the power reducing formula cos 2 ( α ) = cos ( 2 α ) 2 + 1 2 \cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2} cos 2 ( α ) = 2 c o s ( 2 α ) + 2 1 with α = u \alpha= u α = u :
∫ cos 2 ( u ) d u = ∫ ( cos ( 2 u ) 2 + 1 2 ) d u {\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}}} ∫ c o s 2 ( u ) d u = ∫ ( 2 c o s ( 2 u ) + 2 1 ) d u
Apply the constant multiple rule ∫ c f ( u ) d u = c ∫ f ( u ) d u \int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du ∫ c f ( u ) d u = c ∫ f ( u ) d u with c = 1 2 c=\frac{1}{2} c = 2 1 and f ( u ) = cos ( 2 u ) + 1 f{\left(u \right)} = \cos{\left(2 u \right)} + 1 f ( u ) = cos ( 2 u ) + 1 :
∫ ( cos ( 2 u ) 2 + 1 2 ) d u = ( ∫ ( cos ( 2 u ) + 1 ) d u 2 ) {\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)}} ∫ ( 2 c o s ( 2 u ) + 2 1 ) d u = ( 2 ∫ ( c o s ( 2 u ) + 1 ) d u )
Integrate term by term:
∫ ( cos ( 2 u ) + 1 ) d u 2 = ( ∫ 1 d u + ∫ cos ( 2 u ) d u ) 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} 2 ∫ ( c o s ( 2 u ) + 1 ) d u = 2 ( ∫ 1 d u + ∫ c o s ( 2 u ) d u )
Apply the constant rule ∫ c d u = c u \int c\, du = c u ∫ c d u = c u with c = 1 c=1 c = 1 :
∫ cos ( 2 u ) d u 2 + ∫ 1 d u 2 = ∫ cos ( 2 u ) d u 2 + u 2 \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} 2 ∫ cos ( 2 u ) d u + 2 ∫ 1 d u = 2 ∫ cos ( 2 u ) d u + 2 u
Let v = 2 u v=2 u v = 2 u .
Then d v = ( 2 u ) ′ d u = 2 d u dv=\left(2 u\right)^{\prime }du = 2 du d v = ( 2 u ) ′ d u = 2 d u (steps can be seen » ), and we have that d u = d v 2 du = \frac{dv}{2} d u = 2 d v .
So,
u 2 + ∫ cos ( 2 u ) d u 2 = u 2 + ∫ cos ( v ) 2 d v 2 \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} 2 u + 2 ∫ c o s ( 2 u ) d u = 2 u + 2 ∫ 2 c o s ( v ) d v
Apply the constant multiple rule ∫ c f ( v ) d v = c ∫ f ( v ) d v \int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv ∫ c f ( v ) d v = c ∫ f ( v ) d v with c = 1 2 c=\frac{1}{2} c = 2 1 and f ( v ) = cos ( v ) f{\left(v \right)} = \cos{\left(v \right)} f ( v ) = cos ( v ) :
u 2 + ∫ cos ( v ) 2 d v 2 = u 2 + ( ∫ cos ( v ) d v 2 ) 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} 2 u + 2 ∫ 2 c o s ( v ) d v = 2 u + 2 ( 2 ∫ c o s ( v ) d v )
The integral of the cosine is ∫ cos ( v ) d v = sin ( v ) \int{\cos{\left(v \right)} d v} = \sin{\left(v \right)} ∫ cos ( v ) d v = sin ( v ) :
u 2 + ∫ cos ( v ) d v 4 = u 2 + 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} 2 u + 4 ∫ c o s ( v ) d v = 2 u + 4 s i n ( v )
Recall that v = 2 u v=2 u v = 2 u :
u 2 + sin ( v ) 4 = u 2 + sin ( ( 2 u ) ) 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} 2 u + 4 sin ( v ) = 2 u + 4 sin ( ( 2 u ) )
Recall that u = asin ( x ) u=\operatorname{asin}{\left(x \right)} u = asin ( x ) :
sin ( 2 u ) 4 + u 2 = sin ( 2 asin ( 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} 4 sin ( 2 u ) + 2 u = 4 sin ( 2 asin ( x ) ) + 2 asin ( x )
Тому,
∫ 1 − x 2 d x = sin ( 2 asin ( 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} ∫ 1 − x 2 d x = 4 sin ( 2 asin ( x ) ) + 2 asin ( x )
Using the formulas sin ( 2 asin ( α ) ) = 2 α 1 − α 2 \sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}} sin ( 2 asin ( α ) ) = 2 α 1 − α 2 , sin ( 2 acos ( α ) ) = 2 α 1 − α 2 \sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}} sin ( 2 acos ( α ) ) = 2 α 1 − α 2 , cos ( 2 asin ( α ) ) = 1 − 2 α 2 \cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2} cos ( 2 asin ( α ) ) = 1 − 2 α 2 , cos ( 2 acos ( α ) ) = 2 α 2 − 1 \cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1 cos ( 2 acos ( α ) ) = 2 α 2 − 1 , sinh ( 2 asinh ( α ) ) = 2 α α 2 + 1 \sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1} sinh ( 2 asinh ( α ) ) = 2 α α 2 + 1 , sinh ( 2 acosh ( α ) ) = 2 α α − 1 α + 1 \sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1} sinh ( 2 acosh ( α ) ) = 2 α α − 1 α + 1 , cosh ( 2 asinh ( α ) ) = 2 α 2 + 1 \cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1 cosh ( 2 asinh ( α ) ) = 2 α 2 + 1 , cosh ( 2 acosh ( α ) ) = 2 α 2 − 1 \cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1 cosh ( 2 acosh ( α ) ) = 2 α 2 − 1 , simplify the expression:
∫ 1 − x 2 d x = x 1 − x 2 2 + asin ( x ) 2 \int{\sqrt{1 - x^{2}} d x} = \frac{x \sqrt{1 - x^{2}}}{2} + \frac{\operatorname{asin}{\left(x \right)}}{2} ∫ 1 − x 2 d x = 2 x 1 − x 2 + 2 asin ( x )
Додайте константу інтегрування:
∫ 1 − x 2 d x = x 1 − x 2 2 + 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 ∫ 1 − x 2 d x = 2 x 1 − x 2 + 2 asin ( x ) + C
Answer: ∫ 1 − x 2 d x = x 1 − x 2 2 + 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 ∫ 1 − x 2 d x = 2 x 1 − x 2 + 2 asin ( x ) + C