Integrals — Class 12 Maths Solution

: Let $I = \int\limits_0^{\pi /2} {\sqrt {\sin \phi } {{\cos }^5}\phi d\phi } = \int\limits_0^{\pi /2} {\sqrt {\sin \phi } {{\cos }^4}\phi \cos \phi d\phi }$

$= \int\limits_0^{\pi /2} {\sqrt {\sin \phi } {{\left( {1 - {{\sin }^2}\phi } \right)}^2}\cos \phi } d\phi$

Put $\sin \phi = t$ $\Rightarrow$ $\cos \phi d\phi = dt$

When $\phi = 0,t = 0$ and when $\phi = \cfrac{\pi }{2},t = 1$

$\therefore$ $I = \int\limits_0^1 {\sqrt t {{\left( {1 - {t^2}} \right)}^2}dt} = \int\limits_0^1 {\sqrt t \left( {1 - 2{t^2} + {t^4}} \right)dt}$

$= \int\limits_0^1 {\left( {{t^{1/2}} + {t^{9/2}} - 2{t^{5/2}}} \right)dt}$

$= \left[ {\cfrac{2}{3}{t^{3/2}} + \cfrac{2}{{11}}{t^{11/2}} - \cfrac{4}{7}{t^{7/2}}} \right]_0^1 = \cfrac{2}{3} + \cfrac{2}{{11}} - \cfrac{4}{7} = \cfrac{{154 + 42 - 132}}{{3 \times 11 \times 7}} = \cfrac{{64}}{{231}}$