The value of $\int_0^{\pi /2} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \,dx$ is

$e^{\int_0^{\pi / 2} \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x}=$

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 $3 I^2=$

The value of $\int_{-1}^1 \sin^5 x \cos^4 x \, dx$ is

If $\int_{ - a}^a {\sqrt {\frac{{a - x}}{{a + x}}} \,dx = k\pi ,} $ then $k = $

If $\int_{0}^{1} 4 \cot^{-1}(1-x+x^{2}) dx = a \tan^{-1}(2) - b \log_{e}(5)$, where $a, b \in N$, then $(2a+b)$ is equal to :