How do you integrate #intsqrt(1-4x^2)# by trigonometric substitution?

Question

Ezra Adams · Accepted Answer

#sin^2(theta)/2+c#

As shown in the picture below our equation matches the "#sin#" trigonometric substitution.

Therefor,

#x=sin(theta)/2#

#sqrt(1-4x^2)=cos(theta)#

#dx=(2cos(theta))/4 d theta#=#cos(theta)/4#

#intcos(theta)(cos(theta))/2#=#1/2intcos^2(theta)d theta#

#1/2sin^2(theta)+c# or #sin^2(theta)/2+c#

Good Luck!!!

Isaiah Allen · Answer

undefined To integrate $\int \sqrt{1 - 4x^2} \,dx$ using trigonometric substitution, follow these steps:

1. Recognize that the integral involves a square root of a difference of squares, which suggests using a trigonometric substitution.

2. Let $x = \frac{1}{2}\sin(\theta)$, which implies $dx = \frac{1}{2}\cos(\theta) \,d\theta$.

3. Substitute $x$ and $dx$ into the integral and rewrite it in terms of $\theta$:
   $$\int \sqrt{1 - 4\left(\frac{1}{2}\sin(\theta)\right)^2} \cdot \frac{1}{2}\cos(\theta) \,d\theta$$

4. Simplify the expression under the square root:
   $$\sqrt{1 - 4\left(\frac{1}{2}\sin(\theta)\right)^2} = \sqrt{1 - \sin^2(\theta)}$$

5. Use the Pythagorean identity $\cos^2(\theta) = 1 - \sin^2(\theta)$ to simplify the expression under the square root:
   $$\sqrt{1 - \sin^2(\theta)} = \sqrt{\cos^2(\theta)} = |\cos(\theta)|$$

6. Depending on the range of integration, consider whether $|\cos(\theta)|$ needs to be replaced with $\cos(\theta)$ or $-\cos(\theta)$.

7. Rewrite the integral using the trigonometric substitution:
   $$\int |\cos(\theta)| \cdot \frac{1}{2}\cos(\theta) \,d\theta$$

8. Integrate the resulting expression with respect to $\theta$.

9. Finally, substitute back $x = \frac{1}{2}\sin(\theta)$ to express the solution in terms of $x$.