Question

i need answer this one ​

Answer 1

➤Question :-

$\int \: \frac{ \sin( { \tan}^{ - 1} x) }{1 + {x}^{2} } dx = \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}.$

➤Answer :-

$\boxed{ \int \frac{ \sin( { \tan}^{ - 1} x) }{1 + {x}^{2} } dx = - \cos( { \tan}^{ - 1}x )+\: c} \: \\ \\ \sf \: option \: (b) \: is \: correct$

➤Solution :-

We have ,

$\to \: \int \: \frac{ \sin( { \tan}^{ - 1} x) }{1 + {x}^{2} } dx \: \\ \\ { let \: } \: , \: \: \: { \tan}^{ - 1} x = t \: \: \: ..... eq.(1)\\ \\ \sf \red{ differentiate \: with }\: respect \: to \: x \\ \\ \to \frac{d}{dx} { \tan}^{ - 1} x = \frac{dt}{dx} \\ \\ \: \: \: \boxed{ \because \: \frac{d}{dx} { \tan}^{ - 1} x = \frac{1}{1 + {x}^{2} } } \\ \therefore \: \\ \\ \to \: \frac{1}{1 + {x}^{2} } = \frac{dx}{dt} \\ \\ \to \: \frac{1}{1 + {x}^{2} } \: dx = dt \: .....\: eq.(2) \\ \\ hence \: , \sf it \: will \: be \: \\ \\ \to \: \int \: \sin( { \tan}^{ - 1}x ) \: \frac{1}{1 + {x}^{2} } \: dx \\ \\ \sf from \: eq.(2) \\ \: \: \\ \to \: \int \sin(t) \: dt \\ \\ \to \: - \cos(t) + c \\ \\ \sf now \: from \: eq.(1) \\ \\ \to \boxed{ - \cos( { \tan}^{ - 1}x ) \: + c } \\ \\ \to \boxed{ \int \frac{ \sin( { \tan}^{ - 1} x) }{1 + {x}^{2} } dx = - \cos( { \tan}^{ - 1}x )+ \: c}$

Option (b) is correct