Question

$I=\int \frac{\sin x}{\sqrt{1+3{\cos }^{2}x}}dx.$

Answer 1

$I=\int \frac{\sin xdx}{\sqrt{1+3{\cos }^{2}x}}\cos x=t-\sin dx=dtI=-\int \frac{dt}{\sqrt{1+3t^2}}=\frac{-1}{\sqrt{3}}\int \frac{dt}{\sqrt{{\left(\frac{1}{3}\right)}^2+t^2}}=-\frac{1}{\sqrt{3}}\ln |t+\sqrt{\frac{1}{3}+t^2}|+C=-\frac{1}{\sqrt{3}}\ln |\cos x+\sqrt{\frac{1}{3}+{cos}^{2}x}|+C$