If $M=\int_0^{\infty} \frac{\log t}{1+t^3} d t$ and $N=\int_{-\infty}^{\infty} \frac{t e^{2 t}}{1+e^{3 t}} d t$,then

If the value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^2 \cos x}{1+\pi^x}+\frac{1+\sin ^2 x}{1+e^{\sin x^{323}}}\right) d x=\frac{\pi}{4}(\pi+a)-2$,then the value of $a$ is

$\int_5^9 \frac{\log_3 x^2}{\log_3 x^2 + \log_3(588 - 84x + 3x^2)} dx =$

Read the following mathematical statements carefully:
$I.$ $A$ differentiable function $f$ with maximum at $x = c$ $\implies f''(c) < 0$.
$II.$ Antiderivative of a periodic function is also a periodic function.
$III.$ If $f$ has a period $T$ then for any $a \in R$,$\int\limits_0^T {f(x)\,dx} = \int\limits_0^T {f(x + a)\,dx}$.
$IV.$ If $f(x)$ has a maxima at $x = c$,then $f$ is increasing in $(c - h, c)$ and decreasing in $(c, c + h)$ as $h \to 0$ for $h > 0$. Now indicate the correct alternative.

If $I_n = \int_0^{\pi / 4} \tan^n x \, dx$,then $\frac{1}{I_2 + I_4} + \frac{1}{I_3 + I_5} + \frac{1}{I_4 + I_6} = $

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot x}} d x=$