The Correct Answer is:\(I = \frac{\pi}{4}\)
Let \(I = \int^{\frac{π}{2}}_0 \frac{sin^{\frac{3}{2}}xdx}{sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x} ...(1)\)
⇒\(I = \int^{\frac{π}{2}}_0 \frac{sin^{\frac{3}{2}}(\frac{\pi}{2} - x)}{sin^{\frac{3}{2}}(\frac{\pi}{2} - x)+cos^{\frac{3}{2}}(\frac{\pi}{2} - x)} dx \,\,\,\,\,\,\, (\int^a_0 f(x)dx= \int^a_0 f(a-x)dx)\)
⇒\(I = \int^{\frac{π}{2}}_0 \frac{cos^{\frac{3}{2}}x}{sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x}dx...(2)\)
Adding(1)and(2),we obtain
\(2I = \int^{\frac{π}{2}}_0 \frac{sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x}{sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x}dx\)
⇒\(2I = \int^{\frac{π}{2}}_0 1.dx\)
⇒\(2I = [x]^\frac{\pi}{2}_0\)
⇒\(2I = \frac{\pi}{2}\)
⇒\(I = \frac{\pi}{4}\)
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R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
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Find the area of the region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\)