1. ## Integration by substitution

∫√(1-x^2 ) dx

Letting x = sin θ,
dx/dθ = cos θ

∫√(1-x^2 ) dx
=∫√(1-sin^2⁡θ ) cos θ dθ
=∫cos^2 θ dθ
=∫ (1-sin^2⁡θ) dθ
=θ -2sinθ(-cosθ) + c
=θ +2sinθcosθ + c
=sin^(-1)⁡θ + 2x√(1-x^2 ) + c

Is this correct? Please help me to check whether it is correct or not, and point out my mistake, if any. Thank you.

2. ## Re: Integration by substitution

Originally Posted by alexander9408

=∫√(1-sin^2⁡θ ) cos θ dθ
=∫cos^2 θ dθ
Hint: From here, do it by integration by parts twice.

-Dan

3. ## Re: Integration by substitution

Hello, alexander9408!

$\displaystyle\int\sqrt{1-x^2}\,dx$

$\text{Let }\,x = \sin\theta \quad\Rightarrow\quad dx = \cos\theta\,d\theta$

$\displaystyle\text{Substitute: }\:\int\cos^2\!\theta\,d\theta$

Use the identity: . $\cos^2x \:=\:\frac{1+\cos2x}{2}$

$\displaystyle\text{We have: }\:\int \left(\frac{1+\cos2\theta}{2}\right)d\theta \;=\;\tfrac{1}{2}\int\left(1 + \cos2\theta\right)d\theta$

. . . . . $=\;\tfrac{1}{2}\left(\theta + \tfrac{1}{2}\sin2\theta\right) + C \;=\;\tfrac{1}{2}\left(\theta + \tfrac{1}{2}\cdot 2\sin\theta\cos\theta\right) + C$

. . . . . $=\;\tfrac{1}{2}\left(\theta + \sin\theta\cos\theta\right) + C$

Back-substitute: . $\tfrac{1}{2}\left(\arcsin x + x\sqrt{1-x^2}\right) + C$