Question

The value of \(\int\limits^{\frac{1}{2}}_{-\frac{1}{2}}\cos^{-1}x\ dx\) is

• A. π
• B. \(\frac{\pi}{2}\)
• C. 1
• D. \(\frac{\pi^2}{2}\)

Answer 1

The Correct Option is B

$$ I = \int_{-1/2}^{1/2} \cos^{-1} x \, dx. $$

We can use integration by parts. Let $ u = \cos^{-1} x $ and $ dv = dx $, so $ du = \frac{-1}{\sqrt{1-x^2}} dx $ and $ v = x $. Using the integration by parts formula $ \int u \, dv = uv - \int v \, du $, we have:

$$ I = x \cos^{-1} x \Big|_{-1/2}^{1/2} - \int_{-1/2}^{1/2} x \cdot \frac{-1}{\sqrt{1-x^2}} dx. $$

Simplify the second term:

$$ I = x \cos^{-1} x \Big|_{-1/2}^{1/2} + \int_{-1/2}^{1/2} \frac{x}{\sqrt{1-x^2}} dx. $$

To evaluate $ \int \frac{x}{\sqrt{1-x^2}} dx $, let $ w = 1-x^2 $, so $ dw = -2x dx $, or equivalently $ x dx = -\frac{1}{2} dw $. Substituting:

$$ \int \frac{x}{\sqrt{1-x^2}} dx = \int \frac{-\frac{1}{2} dw}{\sqrt{w}} = -\frac{1}{2} \int w^{-1/2} dw = -\frac{1}{2} \cdot \frac{w^{1/2}}{1/2} = -\sqrt{w} = -\sqrt{1-x^2}. $$

Thus:

$$ I = x \cos^{-1} x \Big|_{-1/2}^{1/2} - \sqrt{1-x^2} \Big|_{-1/2}^{1/2}. $$

Evaluate the terms:

$$ I = \left( \frac{1}{2} \cos^{-1} \left( \frac{1}{2} \right) - \left( -\frac{1}{2} \cos^{-1} \left( -\frac{1}{2} \right) \right) \right) - \left( \sqrt{1-\left(\frac{1}{2}\right)^2} - \sqrt{1-\left(-\frac{1}{2}\right)^2} \right). $$

Simplify:

$$ I = \frac{1}{2} \cos^{-1} \left( \frac{1}{2} \right) + \frac{1}{2} \cos^{-1} \left( -\frac{1}{2} \right) - \left( \sqrt{1-\frac{1}{4}} - \sqrt{1-\frac{1}{4}} \right). $$

The square root terms cancel out, leaving:

$$ I = \frac{1}{2} \left( \cos^{-1} \left( \frac{1}{2} \right) + \cos^{-1} \left( -\frac{1}{2} \right) \right). $$

Using $ \cos^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{3} $ and $ \cos^{-1} \left( -\frac{1}{2} \right) = \frac{2\pi}{3} $:

$$ I = \frac{1}{2} \left( \frac{\pi}{3} + \frac{2\pi}{3} \right) = \frac{1}{2} \cdot \frac{3\pi}{3} = \frac{\pi}{2}. $$

Answer 2

We need to find the value of \(\int_{-\frac{1}{2}}^{\frac{1}{2}} \cos^{-1} x \, dx\).

Let's integrate by parts. Let \(u = \cos^{-1} x\) and \(dv = dx\). Then \(du = -\frac{1}{\sqrt{1-x^2}} dx\) and \(v = x\).

\(\int \cos^{-1} x \, dx = x \cos^{-1} x - \int x \cdot (-\frac{1}{\sqrt{1-x^2}}) dx\)

\(\int \cos^{-1} x \, dx = x \cos^{-1} x + \int \frac{x}{\sqrt{1-x^2}} dx\)

For the second integral, let \(w = 1-x^2\), then \(dw = -2x \, dx\), so \(x \, dx = -\frac{1}{2} dw\).

\(\int \frac{x}{\sqrt{1-x^2}} dx = \int \frac{-\frac{1}{2}}{\sqrt{w}} dw = -\frac{1}{2} \int w^{-1/2} dw = -\frac{1}{2} \cdot \frac{w^{1/2}}{1/2} = -w^{1/2} = -\sqrt{1-x^2}\)

So, \(\int \cos^{-1} x \, dx = x \cos^{-1} x - \sqrt{1-x^2}\).

Now we need to evaluate the definite integral:

\(\int_{-\frac{1}{2}}^{\frac{1}{2}} \cos^{-1} x \, dx = [x \cos^{-1} x - \sqrt{1-x^2}]_{-\frac{1}{2}}^{\frac{1}{2}}\)

= \(((\frac{1}{2}) \cos^{-1} (\frac{1}{2}) - \sqrt{1-(\frac{1}{2})^2}) - ((-\frac{1}{2}) \cos^{-1} (-\frac{1}{2}) - \sqrt{1-(-\frac{1}{2})^2}))\)

= \((\frac{1}{2} \cdot \frac{\pi}{3} - \sqrt{\frac{3}{4}}) - (-\frac{1}{2} \cdot \frac{2\pi}{3} - \sqrt{\frac{3}{4}})\)

= \((\frac{\pi}{6} - \frac{\sqrt{3}}{2}) - (-\frac{\pi}{3} - \frac{\sqrt{3}}{2})\)

= \(\frac{\pi}{6} - \frac{\sqrt{3}}{2} + \frac{\pi}{3} + \frac{\sqrt{3}}{2}\)

= \(\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}\)

Therefore, the correct option is (B) \(\frac{\pi}{2}\).