Question

Integrate dx/3sinx+4cosx

Answer 1

`\displaystyle\int(\mathrm dx)/(3\sin x+4\cos x)`

A standard approach would be the tangent half-angle substitution:

`t=\tan\frac x2\implies\mathrm dt=\frac12\sec^2\frac x2\,\mathrm dx`

Then

`\sin x=2\sin\frac x2\cos\frac x2\implies\sin x=(2t)/(1+t^2)`

`\cos x=\cos^2\frac x2-\sin^2\frac x2\implies\cos x=(1-t^2)/(1+t^2)`

from which we get

`\mathrm dx=\frac2{1+t^2}\,\mathrm dt`

So the integral becomes

`\displaystyle\int\frac{\frac2{1+t^2}}{(6t)/(1+t^2)+(4(1-t^2))/(1+t^2)}\,\mathrm dt=\int(\mathrm dt)/(3t+2(1-t^2))=-\int(\mathrm dt)/(2t^2-3t-2)`

Rewrite the denominator as

`2t^2-3t-2=(2t+1)(t-2)`

and expand the integrand into its partial fractions:

`\frac1{2t^2-3t-2}=\frac15\left(\frac1{t-2}-\frac2{2t+1}\right)`

We have

`\displaystyle-\frac15\int\frac1{t-2}-\frac2{2t+1}\,\mathrm dt=-\frac15(\ln|t-2|-\ln|2t+1|)+C`

`=\frac15\ln\left|(2t+1)/(t-2)\right|+C`

`=\frac15\ln\left|(2\tan\frac x2+1)/(\tan\frac x2-2)\right|+C`