Question
$\int {\frac{{dx}}{{\sqrt {16 - 9{x^2}} }}}$
Integrals — Class 12 Maths Solution
Step-by-step Solution
Let $I = \int {\frac{{dx}}{{\sqrt {16 - 9{x^2}} }}} = \int {\frac{{dx}}{{\sqrt {{{(4)}^2} - {{(3x)}^2}} }}} dx = \frac{1}{3}{\sin ^{ - 1}}\left( {\frac{{3x}}{4}} \right) + C$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Integrals. Curated by Sachin Sharma. Free for all students.