$\int \frac{\sqrt{1-x^2} \sin ^{-1} x+x}{\sqrt{1-x^2}} d x=$

Let $5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3$,where $x > 0$. Then $18 \int \limits_1^2 f(x) \, dx$ is equal to:

If $\int \frac{\cos (13 x)-\cos (14 x)}{1+2 \cos (9 x)} d x=\frac{\sin (4 x)}{a}-\frac{\sin (5 x)}{b}+c$,then $a^b=$

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

The value of the integral $\int \frac{\sin \theta \cdot \sin 2 \theta \left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta$ is (where $c$ is a constant of integration)

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