Take the integral: ∫1−x2dx For the integrand 1−x2, substitute x=sin⁡(u)anddx=cos⁡(u)du. Then 1−x2=1−sin2(u)=cos⁡(u)andu=sin−1(x): =∫cos2(u)du Write cos2(u) as 12cos⁡(2u)+12: =∫(12cos⁡(2u)+12)du Integrate the sum term by term and factor out constants: =12∫cos⁡(2u)du+12∫1du For the integrand cos⁡(2u), substitute s=2uandds=2du: =14∫cos⁡(s)ds+12∫1du The integral of cos⁡(s) is sin⁡(s): =sin⁡(s)4+12∫1du The integral of 1 is u: =sin⁡(s)4+u2+constant Substitute back for s=2u: =u2+14sin⁡(2u)+constant Apply the double angle formula sin⁡(2u)=2sin⁡(u)cos⁡(u): =u2+12sin⁡(u)cos⁡(u)+constant Express cos(u) in terms of sin⁡(u) using cos2(u)=1−sin2(u): =u2+12sin⁡(u)1−sin2(u)+constant Substitute back for u=sin−1(x):

Get the solution to "f(x)=∫1−x2!dx"

College
Algebra
Linear algebra
Vectors and spaces

Knight_Snape

Answered question

2021-12-16

f(x)= $\int \sqrt{1 - x^{2}} d x$

Answer & Explanation

Jeffrey Jordon

Expert2022-01-02Added 2605 answers

Take the integral:
$\int \sqrt{1 - x^{2}} d x$

For the integrand $\sqrt{1 - x^{2}}$ , substitute $x = \sin (u) and d x = \cos (u) d u$ . Then $\sqrt{1 - x^{2}} = \sqrt{1 - \sin^{2} (u)} = \cos (u) and u = \sin^{- 1} (x) :$
$= \int \cos^{2} (u) d u$

Write $\cos^{2} (u)$ as $\frac{1}{2} \cos (2 u) + \frac{1}{2}$ :
$= \int (\frac{1}{2} \cos (2 u) + \frac{1}{2}) d u$

Integrate the sum term by term and factor out constants:
$= \frac{1}{2} \int \cos (2 u) d u + \frac{1}{2} \int 1 d u$

For the integrand $\cos (2 u)$ , substitute $s = 2 u and d s = 2 d u$ :
$= \frac{1}{4} \int \cos (s) d s + \frac{1}{2} \int 1 d u$

The integral of $\cos (s)$ is $\sin (s)$ :
$= \frac{\sin (s)}{4} + \frac{1}{2} \int 1 d u$

The integral of 1 is u:
$= \frac{\sin (s)}{4} + \frac{u}{2} + c o n s t a n t$

Substitute back for $s = 2 u$ :
$= \frac{u}{2} + \frac{1}{4} \sin (2 u) + c o n s t a n t$

Apply the double angle formula $\sin (2 u) = 2 \sin (u) \cos (u)$ :
$= \frac{u}{2} + \frac{1}{2} \sin (u) \cos (u) + c o n s t a n t$

Express cos(u) in terms of $\sin (u)$ using $\cos^{2} (u) = 1 - \sin^{2} (u) :$
$= \frac{u}{2} + \frac{1}{2} \sin (u) \sqrt{1 - \sin^{2} (u)} + c o n s t a n t$

Substitute back for $u = \sin^{- 1} (x)$ :

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