Question

integral of suare root 1+sinx.dx​

Answer 1

ANSWER

I=∫

1−sinx

dx

=∫

1−sinx×(

1+sinx

1+sinx

)

dx

=∫

(

1+sinx

1

2

−sin

2

x

)

dx

=∫(

1+sinx

cosx

)dx

let 1+sinx=t

then cosx dx=dt

orI=∫(

t

1

)dt

=(

(

2

1

)

t

(

2

1

)

)+C

2

1+sinx

+C