If $f(x) = \int x^2 \cos^2 x (2x \tan^2 x - 2x - 6 \tan x) dx$ and $f(0) = \pi$,then $f(x) =$

Let $F(x)$ be an indefinite integral of $\sin ^2 x$.
$STATEMENT -1$ : The function $F(x)$ satisfies $F(x+\pi)=F(x)$ for all real $x$. because
$STATEMENT -2$: $\sin ^2(x+\pi)=\sin ^2 x$ for all real $x$.

If $\int x^{49} \left[ \operatorname{Tan}^{-1} x^{50} + \frac{x^{50}}{1 + x^{100}} \right] dx = \frac{x^n}{k} f(x) + c$,then $f(x) - f\left(\sqrt[k]{x^n}\right) =$

If an antiderivative of $f(x)$ is $e^x$ and an antiderivative of $g(x)$ is $\cos x$, then $\int f(x) \cos x \, dx + \int g(x) e^x \, dx =$

If $f(x) = \frac{\sin^2 \pi x}{1+\pi^x}$,then $\int (f(x) + f(-x)) \, dx$ is equal to

If $I_n = \int \tan^n x \ dx$,and $I_0 + I_1 + 2 I_2 + 2 I_3 + 2 I_4 + I_5 + I_6 = \sum_{K=1}^n \frac{\tan^K x}{K}$,then $n = $