use the substitution ( t=\tan left(\frac{x}{2}\right) ) to transform the integrand into a rational function…

Explanation:

Step1: Use trigonometric identities

We know that (\sin(2x)=2\sin x\cos x), and with (t = \tan\frac{x}{2}), we have (\sin x=\frac{2t}{1 + t^{2}}), (\cos x=\frac{1 - t^{2}}{1 + t^{2}}), and (dx=\frac{2}{1 + t^{2}}dt). When (x = 0), (t=\tan(0)=0); when (x=\frac{\pi}{2}), (t=\tan(\frac{\pi}{4}) = 1). The integral (\int_{0}^{\frac{\pi}{2}}\frac{7\sin(2x)}{2+\cos x}dx=\int_{0}^{\frac{\pi}{2}}\frac{14\sin x\cos x}{2+\cos x}dx). Substituting the values of (\sin x), (\cos x) and (dx) in terms of (t): [ \begin{align*} \int_{0}^{1}\frac{14\times\frac{2t}{1 + t^{2}}\times\frac{1 - t^{2}}{1 + t^{2}}}{2+\frac{1 - t^{2}}{1 + t^{2}}}\times\frac{2}{1 + t^{2}}dt&=\int_{0}^{1}\frac{14\times2t(1 - t^{2})\times2}{(1 + t^{2})^{2}(2(1 + t^{2})+(1 - t^{2}))}dt\ &=\int_{0}^{1}\frac{56t(1 - t^{2})}{(1 + t^{2})^{2}(2 + 2t^{2}+1 - t^{2})}dt\ &=\int_{0}^{1}\frac{56t(1 - t^{2})}{(1 + t^{2})^{2}(t^{2}+ 3)}dt \end{align*} ] Let (u=t^{2}), then (du = 2tdt). When (t = 0), (u = 0); when (t = 1), (u = 1). The integral becomes (\int_{0}^{1}\frac{28(1 - u)}{(1 + u)^{2}(u + 3)}du). Using partial - fraction decomposition: (\frac{1 - u}{(1 + u)^{2}(u + 3)}=\frac{A}{1 + u}+\frac{B}{(1 + u)^{2}}+\frac{C}{u + 3}) [ \begin{align*} 1-u&=A(1 + u)(u + 3)+B(u + 3)+C(1 + u)^{2}\ \end{align*} ] Let (u=-1), then (1-(-1)=B(-1 + 3)), so (B = 1). Let (u=-3), then (1-(-3)=C(1-3)^{2}), so (C = 1). Comparing the coefficient of (u^{2}): (0=A + C), so (A=-1).

So (\frac{1 - u}{(1 + u)^{2}(u + 3)}=-\frac{1}{1 + u}+\frac{1}{(1 + u)^{2}}+\frac{1}{u + 3})

The integral (\int_{0}^{1}\frac{28(1 - u)}{(1 + u)^{2}(u + 3)}du=28\int_{0}^{1}\left(-\frac{1}{1 + u}+\frac{1}{(1 + u)^{2}}+\frac{1}{u + 3}\right)du)

Step2: Integrate term - by - term

[ \begin{align*} 28\int_{0}^{1}\left(-\frac{1}{1 + u}+\frac{1}{(1 + u)^{2}}+\frac{1}{u + 3}\right)du&=28\left[-\ln(1 + u)-\frac{1}{1 + u}+\ln(u + 3)\right]_{0}^{1}\ &=28\left[\left(-\ln(2)-\frac{1}{2}+\ln(4)\right)-\left(-\ln(1)-1+\ln(3)\right)\right]\ &=28\left(-\ln(2)-\frac{1}{2}+2\ln(2)+1-\ln(3)\right)\ &=28\left(\ln(2)+\frac{1}{2}-\ln(3)\right)\ &=28\ln\frac{2}{3}+14 \end{align*} ]

Answer:

(28\ln\frac{2}{3}+14)