# $I=3\sqrt2\int_{0}^{x}\frac{\sqrt{1+\cos t}}{17-8\cos t}dt$.If $0<x<\pi$ and $\tan I=\frac{2}{\sqrt3}$

Let $I=3\sqrt2\int_{0}^{x}\frac{\sqrt{1+\cos t}}{17-8\cos t}dt$.If $0<x<\pi$ and $\tan I=\frac{2}{\sqrt3}$.Find $x$.

$\int\frac{\sqrt{1+\cos t}}{17-8\cos t}dt=\int\frac{\sqrt2\cos\frac{t}{2}}{17-8\times\frac{1-\tan^2\frac{t}{2}}{1+\tan^2\frac{t}{2}}}dt=\int\frac{\sqrt2\cos\frac{t}{2}(1+\tan^2\frac{t}{2})}{17+17\tan^2\frac{t}{2}-8+8\tan^2\frac{t}{2}}dt=\int\frac{\sqrt2\cos\frac{t}{2}(\sec^2\frac{t}{2})}{9+25\tan^2\frac{t}{2}}dt$

$=\int\frac{\sqrt2(\sec^2\frac{t}{2})}{(\sec\frac{t}{2})(9+25\tan^2\frac{t}{2})}dt$
$=\int\frac{\sqrt2(\sec^2\frac{t}{2})}{\sqrt{1+\tan^2\frac{t}{2}}(9+25\tan^2\frac{t}{2})}dt$
Put $\tan\frac{x}{2}=t$
$=\int\frac{\sqrt2(\sec^2\frac{t}{2})}{\sqrt{1+\tan^2\frac{t}{2}}(9+25\tan^2\frac{t}{2})}dt$
$=\int\frac{\sqrt2 }{2\sqrt{1+p^2}(9+25p^2)}dp$

$$\frac{\sqrt{1+\cos t}}{17-8\cos t}=\dfrac{\sqrt2\left|\cos\dfrac t2\right|}{17-8\left(1-2\sin^2\dfrac t2\right)}$$
Set $\sin\dfrac t2=u$