class 12 maths integrals

$\cfrac{1}{{\sqrt {9 - 25{x^2}} }}$

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📘 Integrals NCERT,ex.7.4,Q.4,Page 315 SA

$\cfrac{1}{{\sqrt {9 - 25{x^2}} }}$

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

: Let$I = \int {\cfrac{{dx}}{{\sqrt {9 - 25{x^2}} }}} = \cfrac{1}{5}\int {\cfrac{{dx}}{{\sqrt {\cfrac{9}{{25}} - {x^2}} }}} = \cfrac{1}{5}\int {\cfrac{{dx}}{{\sqrt {{{\left( {\cfrac{3}{5}} \right)}^2} - {x^2}} }}}$

$= \cfrac{1}{5}{\sin ^{ - 1}}\left( {\cfrac{x}{{3/5}}} \right) + C = \cfrac{1}{5}{\sin ^{ - 1}}\left( {\cfrac{{5x}}{3}} \right) + C$

Alternative solution :
Let $I = \int {\cfrac{{dx}}{{\sqrt {9 - 25{x^2}} }}} = \int {\cfrac{{dx}}{{\sqrt {9 - {{\left( {5x} \right)}^2}} }}}$

Put $5x = 3\sin \theta$ $\Rightarrow$ $5dx = 3\cos \theta \,d\theta$

$\therefore$ $I = \cfrac{1}{5}\int {\cfrac{{3\cos \theta \,d\theta }}{{\sqrt {9 - 9{{\sin }^2}\theta } }}} = \cfrac{1}{5}\int {\cfrac{{\cos \theta \,d\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}}$

$= \cfrac{1}{5}\int {d\theta } = \cfrac{1}{5}\theta + C = \cfrac{1}{5}{\sin ^{ - 1}}\left( {\cfrac{{5x}}{3}} \right) + C$

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