$\int_{0}^{\pi} \cos^3 x \, dx = $

$\int_0^1 \frac{8 \log (1+x)}{1+x^2} \,d x=$

If for non-zero $x,$ $af(x) + bf\left( {\frac{1}{x}} \right) = \frac{1}{x} - 5,$ where $a \ne b,$ then $\int_1^2 {f(x)\,dx = } $

Let $T > 0$ be a fixed number. $f: R \rightarrow R$ is a continuous function such that $f(x+T) = f(x)$ for all $x \in R$. If $I = \int_0^T f(x) dx$,then find the value of $\int_0^{5T} f(2x) dx$.

If $f : R \rightarrow R$ is a continuous function satisfying $\int \limits_0^{\pi / 2} f(\sin 2x) \cdot \sin x \, dx + \alpha \int \limits_0^{\pi / 4} f(\cos 2x) \cdot \cos x \, dx = 0$,then $\alpha$ is equal to

$\int_0^{\pi / 2} \log _e(\sin 2 x) d x$