The area of the region bounded by the curve y = \(\frac{1}{1 + \sqrt{\tan {x}}}\) and the x-axis between the ordinates x = \(\frac{\pi}{6}\) and x = \(\frac{\pi}{3}\) is :
Answer & explanation
Correct answer: option 4
y = \(\frac{1}{1 + \sqrt{\tan {x}}}\)
= \(\frac{\sqrt{\cos {x}}}{\sqrt{\ sin {x}} + \sqrt{\cos {x}}}\)
Area bounded between the intervals x = \(\frac{\pi}{6}\) and x = \(\frac{\pi}{3}\) is given by
A = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}d\)x [\(\frac{\sqrt{\cos {x}}}{\sqrt{\ sin {x}} + \sqrt{\cos {x}}}\)]
= \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}d\)x [\(\frac{\sqrt{\cos {\frac{\pi}{3} + \frac{\pi}{6} -x}}}{\sqrt{\ sin {\frac{\pi}{3} + \frac{\pi}{6} -x}} + \sqrt{\cos {\frac{\pi}{3} + \frac{\pi}{6} -x}}}\)]
= \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}d\)x [\(\frac{\sqrt{\sin {x}}}{\sqrt{\ sin {x}} + \sqrt{\cos {x}}}\)]
2A = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}d\)x
⇒ A = \(\frac{1}{2} [\frac{\pi}{3} - \frac{\pi}{6}] \)
= \(\frac{\pi}{12}\) sq. units.