first write

sin 3x = 3 sin x - 4 sin³ x

so integral becomes

= 1/ 4 sinx - 4sin ³ x

take 4 sin x common from denom

= 1 / 4sin x ( 1 - sin² x )

= 1 / 4 sinx X cos ² x

= sec²x / 4 sin x

= 1 + tan²x / 4 sin x

= 1/4 [ 1 / sinx + tan²x / sin x ]

= 1/4 [ cosec x + sinx / cos² x ]

= 1/4 [ cosec x + sec x X tanx ]

so the ans is

1/4 [ log ( cosec x - cot x ) + sec x + C ]

He has simply integrated the 2nd last step

Quite an easy integral.

sinx+sin3x= 4(sinx)(2-cos²x)

Multiply numerator and denominator by sinx.

The denominator becomes 4(sin²x)(2-cos²x)= 4(1-cos²x)(2-cos²x)  [note that you have sinx dx in the numerator]

Now take cosx=t

The denominator becomes (1-t²)(2-t²) [the numerator is simply dt]

write the numerator as (2-t²)-(1-t²)

Now, I hope, you can do the rest.

Rate me.

sin3x+sinx=4sinxcos^2x

sec^2x/sinx

now integrate it by using integration by parts

by taking sec^2x as second function and 1/sinx as first function

then u get

1/4sinx+integral tanx/sin^x

integral cosec2x/2

integral1+tan^2x/2tanx

1/2integralsec^2x/2tanx

tanx=t

d/dxtanx=sec^2x

integral dt/t

1/4(log(tanx)

final answer is tanx/4sinx+1/4 log(tanx)

plz correct me if i am wrong