$\smallint \,\,\frac{{\cos \,\,x - 1}}{{\sin \,\,x + 1}}\,\,{e^x}\,dx$ is equal to

$c\,\, - \,\,\frac{{{e^x}\,\,\sin \,\,x}}{{1 + \,\,\sin \,\,x}}$

$c\,\, - \,\,\frac{{{e^x}}}{{1 + \,\,\sin \,\,x}}$

$c\,\, - \,\,\frac{{{e^x}\,\,\cos \,\,x}}{{1 + \,\,\sin \,\,x}}$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \,\,\frac{{\cos \,\,x - 1}}{{\sin \,\,x + 1}}\,\,{e^x}\,dx

is equal to

View answer and explanation

Correct option: A

\frac{{{e^x}\,\,\cos \,\,x}}{{1\,\, + \,\,\sin \,\,x}}\,\, + \,\,c

Option A (

\frac{{{{{e^x}}\cos \\,x}}}}{{{{1\\,\\ + \\,\sin \\,x}}}}\\ + \\,c

) is correct because the integral can be solved by recognizing a pattern similar to

\int e^x [f(x) + f'(x)] dx = e^x f(x) + C