An integral misses tangent

Let $I=\int_0^{\frac{\pi}{2}}\frac{\sin ^2x}{3+2\sin x+2\cos x}dx$ and $x=\frac{\pi}{2}-u$ $I=\int_{\frac{\pi}{2}}^0\frac{\sin ^2(\frac{\pi}{2}-u)}{3+2\sin (\frac{\pi}{2}-u)+2\cos (\frac{\pi}{2}-u)}(-du)=\int_0^{\frac{\pi}{2}}\frac{\cos ^2u}{3+2\cos u+2\sin u}du=\int_0^{\frac{\pi}{2}}\frac{\cos ^2x}{3+2\sin x+2\cos x}dx$ $2I=\int_0^{\frac{\pi}{2}}\frac{\sin ^2x}{3+2\sin x+2\cos x}dx+\int_0^{\frac{\pi}{2}}\frac{\cos ^2x}{3+2\sin x+2\cos x}dx=\int_0^{\frac{\pi}{2}}\frac{\sin ^2x+\cos ^2x}{3+2\sin x+2\cos x}dx=\int_0^{\frac{\pi}{2}}\frac{1}{3+2\sin x+2\cos x}dx$ Now let $t=\tan{\frac{x}{2}}$ , $dx=\frac{2dt}{1+t^2}$ , $\sin x=\frac{2t}{1+t^2}$ , $\cos x=\frac{1-t^2}{1+t^2}$ $2I=\int_0^1\frac{1}{3+2(\frac{2t}{1+t^2})+2(\frac{1-t^2}{1+t^2})}(\frac{2dt}{1+t^2})$ $I=\int_0^1\frac{1}{3(1+t^2)+4t+2(1-t^2)}dt=\int_0^1\frac{1}{t^2+4t+5}dt=\int_0^1\frac{1}{(t+2)^2+1}dt$ Let $t+2=\tan \theta$ , $dt=\sec ^2\theta \, d\theta$ $I=\int_{\tan ^{-1}2}^{\tan ^{-1}3}\frac{\sec ^2\theta \, d\theta}{\tan ^2\theta +1}=\int_{\tan ^{-1}2}^{\tan ^{-1}3}1\, d\theta$ $I=\tan ^{-1}3-\tan ^{-1}2=\tan ^{-1}(\tan({\tan ^{-1}3-\tan ^{-1}2}))=\tan ^{-1}\frac{\tan (\tan ^{-1}3)-\tan (\tan ^{-1}2)}{1+(\tan (\tan ^{-1}3))(\tan (\tan ^{-1}2))}=\tan ^{-1}\frac{3-2}{1+(3)(2)}$ Therefore, $I=\tan ^{-1}\frac{1}{k}=\tan ^{-1}\frac{1}{7}$ So $k=\boxed{7}$

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac {\sin^2 x}{3+2\sin x+2\cos x} dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac {\sin^2 x}{3+2\sin x+2\cos x}+\frac {\sin^2 \left(\frac \pi 2- x\right)}{3+2\sin \left(\frac \pi 2- x\right)+2\cos \left(\frac \pi 2- x\right)}\right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac {\sin^2 x}{3+2\sin x+2\cos x}+\frac {\cos^2 x}{3+2\cos x + 2\sin x}\right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac 1{3+2\sin x+2\cos x} dx & \small \color{#3D99F6} \text{Let }t = \tan \frac x2 \implies dt = \frac 12 \sec^2 \frac x2 \ dx \\ & = \frac 12 \int_0^1 \frac 1{3+\frac {4t}{1+t^2}+\frac {2-2t^2}{1+t^2}} \times \frac 2{1+t^2} dt & \small \color{#3D99F6} \implies \sin x = \frac {2t}{1+t^2} \implies \cos x = \frac {1-t^2}{1+t^2} \\ & = \int_0^1 \frac 1{t^2+4t+5} dt \\ & = \int_0^1 \frac 1{(t+2)^2+1} dt \\ & = \arctan (t+2) \bigg|_0^1 \\ & = \arctan 3 - \arctan 2 \\ & = \arctan \left(\frac {3-2}{1+3\cdot 2}\right) \\ & = \arctan \frac 17 \end{aligned}$

Therefore, $k = \boxed 7$ .