Evaluate the definite integral ∫_0^(1/2) dx/√(1-4x^2).

Evaluate the definite integral $\int_{0}^{\frac{1}{2}} \frac{d x}{1 - 4 x ^{2}}$ .

hard 4 marks Class 12 · Mathematics · Inverse Trigonometric Functions

Answer

\frac{π}{4}

The correct answer is $\frac{π}{4}$ — Evaluate: $\frac{1}{2} [sin^{- 1} u]_{0}^{1} = \frac{1}{2} (sin^{- 1} 1 - sin^{- 1} 0) = \frac{1}{2} (\frac{π}{2} - 0) = \frac{π}{4} .$

Step-by-step solution

Factor the constant inside the square root: $1 - 4 x^{2} = 1 - (2 x)^{2}$ .
Make the substitution $u = 2 x$ so that $d u = 2 d x$ or $d x = \frac{d u}{2}$ .
When $x = 0, u = 0;$ when $x = \frac{1}{2}, u = 1$ .
Rewrite the integral: $\int_{0}^{\frac{1}{2}} \frac{d x}{1 - 4 x ^{2}} = \int_{0}^{1} \frac{\frac{d u}{2}}{1 - u ^{2}} = \frac{1}{2} \int_{0}^{1} \frac{d u}{1 - u ^{2}} .$
The antiderivative of $\frac{1}{1 - u ^{2}}$ is $sin^{- 1} u$ .
Evaluate: $\frac{1}{2} [sin^{- 1} u]_{0}^{1} = \frac{1}{2} (sin^{- 1} 1 - sin^{- 1} 0) = \frac{1}{2} (\frac{π}{2} - 0) = \frac{π}{4} .$

Common mistake: Forgetting the factor

\frac{1}{2}

after the substitution, which leads to

\frac{π}{2}

instead of

\frac{π}{4}

Concept tested: Integration leading to inverse sine

Published 2 September 2026 · How this solution was prepared and checked

Evaluate the definite integral ∫021​​1−4x2​dx​.

Step-by-step solution

Related questions

Evaluate the definite integral $\int_{0}^{\frac{1}{2}} \frac{d x}{1 - 4 x ^{2}}$ .