Question

Let F(x) be the antiderivative of $f\left(x\right)=1/(3+5sinx+3cosx)$ whose graph passes through the point (0, 0). Then $F\left(\pi /2\right)-\frac{1}{5}\log \frac{8}{3}+1982$ is equal to

Answer 1

$F\left(x\right)=\int f\left(x\right)dx=\int \frac{1}{3+5\sin x+3\cos x}dx$
Substituting $u=\tan \frac{x}{2}\Rightarrow du=\frac{1}{2}{\sec }^2\frac{x}{2}dx$
$\sin x=\frac{2u}{u^2+1},\cos x=\frac{1-u^2}{u^2+1},dx=\frac{2du}{1-u^2}F\left(x\right)=\int \frac{1}{5u+3}du=\frac{\log (5u+3)}{5}+c=\frac{\log (5\tan \frac{x}{2}+3)}{5}+c$
As is passes through $(0,0)$
$F\left(0\right)=0\Rightarrow \frac{\log \left(3\right)}{5}+c=0\Rightarrow c=-\frac{\log \left(3\right)}{5}$
Then
$F\left(\frac{\pi }{2}\right)=\frac{\log (5\tan \frac{\pi }{4}+3)}{5}-\frac{\log \left(3\right)}{5}=\frac{1}{5}\log \frac{8}{3}$
Therefore, $F\left(\frac{\pi }{2}\right)-\frac{1}{5}\log \frac{8}{3}+1982=1982$