Integrate the following function w.r.t.x
`(i)f(x)=sinxcos3x`
`(ii) f(x)=sin^(3)x`
`(iii)f(x)=(sinx)/(sin(x-alpha))`

Explanation

`(i)` We have ,
`l=intsinxcos3xdx`
`impliesl=(1)/(2)int2sinxcos3xdx`
`impliesl=(1)/(2)int(sin4x-sin2x)dx` `{2sinAcosB=sin(A+B)+sin(A-B)}`.
`:.l=(-1)/(8)cos4x+(1)/(4)cos2x+C`
`(ii)` We have,
`l=intsin^(3)xdx`
Using identify,
`sin3x=3sinx-4sin^(3)x`
`impliessin^(3)x=(3sinx-sin3x)/(4)`
`:.l=(1)/(4)int{3sinx-sin3x}dx`
`:.l=(-3)/(4)cosx+(1)/(12)cos3x+C`
`(iii)` We have,
`l=int(sinx)/(sin(x-alpha))dx`
We can write,
`I=int(sin{(x-alpha)+alpha})/(sin(x-alpha))dx`
Using `sin(A+B)=sinAcosB+cosAsinB`
We get,
`l=int(sin(x-alpha)cosalpha+cos(x-alpha)sinalpha)/(sin(x-alpha))dx`
`:.l=intcosalphadx+int(sinalpha)/(tan(x-alpha))dx`
`:.l=cosalphaintdx+sinalphaintcot(x-alpha)dx`
`impliesl=xcosalpha+sinalpha log|sin(x-alpha)|+C`