Question

$\displaystyle \tt \red{ \int^{}_{} \tan \bigg( \sin ^{ - 1} ( {e}^{x} ) \bigg) \: dx}$​

Answer 1

Answer:

$\huge \green{ \sin^{ - 1}( {e}^{x} ) + c}$

Step-by-step explanation:

$\red{convert \: sin {}^{ - 1} \: to \: \: tan {}^{ - 1} \: \: using } \\ \red{\: \: trignomatric \: formulas \: }$

$\int^{}_{} \tan \bigg( \sin ^{ - 1} ( {e}^{x} ) \bigg) \: dx \\ \\ = \int^{}_{} \tan \bigg( \tan ^{ - 1} ( {{e}^{x} \over{ \sqrt{1 - {e}^{2x}} }} ) \bigg) \: dx \\ \\ = \int \: \frac{ {e}^{x} }{ \sqrt{1 - {e}^{2x} } } \: dx \\ \\ put \: \ {e}^{x} = y \implies \: {e}^{x} \: dx = dy \\ \\ = \int \: \frac{dy}{ \sqrt{1 - {y}^{2} } } \\ \\ = \sin {}^{ - 1} (y) + c \\ \\ = \sin {}^{ - 1} ( {e}^{x} ) + c$