Question

Integrate dx/3sinx+4cosx

Answer 1

$\displaystyle\int\frac{\mathrm dx}{3\sin x+4\cos x}$

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

$t=\tan\dfrac x2\implies\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx$

Then

$\sin x=2\sin\dfrac x2\cos\dfrac x2\implies\sin x=\dfrac{2t}{1+t^2}$

$\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2\implies\cos x=\dfrac{1-t^2}{1+t^2}$

from which we get

$\mathrm dx=\dfrac2{1+t^2}\,\mathrm dt$

So the integral becomes

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

Rewrite the denominator as

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

and expand the integrand into its partial fractions:

$\dfrac1{2t^2-3t-2}=\dfrac15\left(\dfrac1{t-2}-\dfrac2{2t+1}\right)$

We have

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

$=\dfrac15\ln\left|\dfrac{2t+1}{t-2}\right|+C$

$=\dfrac15\ln\left|\dfrac{2\tan\frac x2+1}{\tan\frac x2-2}\right|+C$