class 12 maths integrals

$\int\limits_{}^{\pi /2} {\cfrac{{{{\sin }^{3/2}}xdx}}{{{{\sin }^{3/2}}x + {{\cos }^{3/2}}x}}}$

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📘 Integrals NCERT,ex.7.11,Q.3,Page 347 SA

$\int\limits_{}^{\pi /2} {\cfrac{{{{\sin }^{3/2}}xdx}}{{{{\sin }^{3/2}}x + {{\cos }^{3/2}}x}}}$

Official Solution

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Let $I = \int\limits_0^{\pi /2} {\cfrac{{{{\sin }^{3/2}}x}}{{{{\sin }^{3/2}}x + {{\cos }^{3/2}}x}}} dx$

…(i)
$\Rightarrow$ $I = \int\limits_0^{\pi /2} {\cfrac{{{{\sin }^{3/2}}\left( {\cfrac{\pi }{2} - x} \right)}}{{{{\sin }^{3/2}}\left( {\cfrac{\pi }{2} - x} \right) + {{\cos }^{3/2}}\left( {\cfrac{\pi }{2} - x} \right)}}} dx$

$= \int\limits_0^{\pi /2} {\cfrac{{{{\cos }^{3/2}}x}}{{{{\sin }^{3/2}}x + {{\cos }^{3/2}}x}}dx}$

…(ii)
Adding (i) and (ii), we have
$2I = \int\limits_0^{\pi /2} {\cfrac{{si{n^{3/2}}x + {{\cos }^{3/2}}x}}{{{{\sin }^{3/2}}x + {{\cos }^{3/2}}x}}dx}$

$= \int\limits_0^{\pi /2} {1dx} = \left[ x \right]_0^{\pi /2} = \cfrac{\pi }{2} - 0 = \cfrac{\pi }{2}$

$\therefore$ $I = \cfrac{\pi }{4}$

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