Question

Integration of dx/(3sinx+ 4cosx)?

Answer 1

`intdx/(3sinx+4cosx)=-1/5ln|csc(x+alpha)+cot(x+alpha)|+c`, where `alpha=tan^(-1)(4/3)`

Explanation:

`3sinx+4cosx`

= `5(sinx xx 3/5+cosx xx 4/5)`

= `5sin(x+alpha)`, where `tanalpha=4/3` or `alpha=tan^(-1)(4/3)`

Hence, `intdx/(3sinx+4cosx)`

= `1/5intcsc(x+alpha)dx`

Now let `x+alpha=u` then `dx=du` and

`1/5intcsc(x+alpha)dx=1/5intcscudu`

= `1/5int(cscu(cscu+cotu))/(cscu+cotu)du`

= `-1/5int(-csc^2u-cscucotu)/(cscu+cotu)du`

if `cscu+cotu=v`, then `dv=(-csc^2u-cscucotu)du`

and our integral becomes

`-1/5int(dv)/v=-1/5ln|v|+c`

= `-1/5ln|csc(x+alpha)+cot(x+alpha)|+c`, where `alpha=tan^(-1)(4/3)`