Question

∫sqrt(1-4x^2)dx

Answer 1

\frac{x}{2}\sqrt{1-4x^2} + \frac{1}{4}\sin^{-1}(2x) + C

Answer 2

To evaluate the integral $\int \sqrt{1-4x^2}dx$, we can use a substitution method to transform it into a standard integral form.

1. Rewrite the integrand: The expression inside the square root can be written as $1 - (2x)^2$. So, the integral becomes $\int \sqrt{1-(2x)^2}dx$.

2. Perform substitution: Let $u = 2x$. Differentiating both sides with respect to $x$, we get: $du = 2dx$ From this, we can express $dx$ as: $dx = \frac{1}{2}du$

3. Substitute into the integral: Substitute $u$ and $dx$ into the integral: $\int \sqrt{1-u^2} \left(\frac{1}{2}du\right) = \frac{1}{2} \int \sqrt{1-u^2}du$

4. Apply the standard integral formula: This integral is now in the standard form $\int \sqrt{a^2-x^2}dx$, where $a=1$ and the variable is $u$. The standard formula for this type of integral is: $\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$ Applying this formula with $a=1$ and variable $u$: $\frac{1}{2} \left[ \frac{u}{2}\sqrt{1-u^2} + \frac{1^2}{2}\sin^{-1}\left(\frac{u}{1}\right) \right] + C$ $= \frac{1}{2} \left[ \frac{u}{2}\sqrt{1-u^2} + \frac{1}{2}\sin^{-1}(u) \right] + C$

5. Substitute back the original variable: Now, substitute $u=2x$ back into the expression: $\frac{1}{2} \left[ \frac{2x}{2}\sqrt{1-(2x)^2} + \frac{1}{2}\sin^{-1}(2x) \right] + C$ $= \frac{1}{2} \left[ x\sqrt{1-4x^2} + \frac{1}{2}\sin^{-1}(2x) \right] + C$ Distribute the $\frac{1}{2}$: $= \frac{x}{2}\sqrt{1-4x^2} + \frac{1}{4}\sin^{-1}(2x) + C$

The final answer is $\frac{x}{2}\sqrt{1-4x^2} + \frac{1}{4}\sin^{-1}(2x) + C$.

Explanation of the solution:

The integral $\int \sqrt{1-4x^2}dx$ is transformed into a standard form using the substitution $u=2x$, which implies $dx = \frac{1}{2}du$. This yields $\frac{1}{2}\int \sqrt{1-u^2}du$. Applying the standard integral formula $\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$ with $a=1$ and substituting back $u=2x$ gives the result.