$\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$

Let $f: [2, 5] \to [2, 5]$ be a bijective function such that $\frac{d}{dx}(f^{-1}(x)) > 0$ for all $x \in [2, 5]$. Then $\int_{2}^{5} (f(x) + f^{-1}(x)) dx$ is

Let $h(x) = \int\limits_0^x {g(t)dt}$,where $g(x)$ is a differentiable and odd function $\forall x \in R$ and $g(x)$ is periodic with period $3$.
Statement $1: h(x) + h(-x) = 0$ $\forall x \in R$
Statement $2: h(x) + h(-x) = 2 \int\limits_0^x {g(t)dt}$ $\forall x \in R$
Statement $3: h(3n) = 0$ $\forall n \in I$
Then which of the following statement$(s)$ is/are true?

For $n \in N$,if $I_n = \int \frac{\sin nx}{\sin x} dx = \frac{2}{n-1} \sin(n-1)x + I_{n-2}$ and $\int_0^\pi \frac{\sin nx}{\sin x} dx = \frac{k\pi}{2}$,then $k =$

The value of $\int_{-2}^{2} (ax^3 + bx + c) dx$ depends on

Let $I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. Given that $\int \frac{d x}{1+k x^2}=\frac{1}{\sqrt{k}} \tan ^{-1}(\sqrt{k} x)+c, \tan ^{-1}(0)=0$ and $\tan ^{-1}(\sqrt{3})=\frac{\pi}{3}$. Then $3 I^2=$