Let I = - 1 x dx Put - 1 x = t dx = dt I = t dt = 2 t 2 + C = 2 ( - 1 x ) 2 + C

Integrals — Class 12 Maths Solution

ncert exercise SA NCERT,ex.7.2,Q.23,Page 305

Question

$\cfrac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}$

Step-by-step Solution

Let$I = \int {\cfrac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx}$

Put ${\sin ^{ - 1}}x = t$ $\Rightarrow$ $\cfrac{1}{{\sqrt {1 - {x^2}} }}dx = dt$

$\Rightarrow$ $I = \int t dt = \cfrac{1}{2}{t^2} + C = \cfrac{1}{2}{\left( {{{\sin }^{ - 1}}x} \right)^2} + C$

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Integrals. Curated by Sachin Sharma. Free for all students.