Evaluate the following integral ∫1/(cosx (〖sin〗^2 x+1) ) dx
Put z=sinx , cosz=√(1-z^2 ) , dx=1/(√(1-z^2 ) ) dzI=∫〖1/(√(1-z^2 ) (z^2+1) ) dz/(√(1-z^2 ) )〗=∫dz/(1-z^2 )(1+z^2 ) Using partial fractions:I=1/2 ∫dz/((1-z^2 ) )+1/2 ∫dz/((1+z^2 ) )=1/2 tanh〖x^(-1) 〗+1/2 tan^(-1)z+c
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Put z=sinx , cosz=√(1-z^2 ) , dx=1/(√(1-z^2 ) ) dz
need an explanation for this answer? contact us directly to get an explanation for this answerI=∫〖1/(√(1-z^2 ) (z^2+1) ) dz/(√(1-z^2 ) )〗=∫dz/(1-z^2 )(1+z^2 )
Using partial fractions:
I=1/2 ∫dz/((1-z^2 ) )+1/2 ∫dz/((1+z^2 ) )=1/2 tanh〖x^(-1) 〗+1/2 tan^(-1)z+c