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

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Answer 1 · 2017-09-01T14:10:44

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

$3sinx+4cosx$

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

= $5sin(x+alpha)$ , 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 $alpha=tan^(-1)(4/3)$