Help with this Trigonometry integral I've to find this integral:
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx$$
So far as I know to go:
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(4^\pi\cdot \frac{1}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(4^\pi \cot(x)+\tanh^{-1}(x)-4\sec^2( x)\right)dx$$
Thanks a lot!

Hint. Each term of the integrand has an elementary antiderivative on an appropriate domain.


*

*We have


$$
\int\frac1{\tan x}\:dx=\int\frac{\cos x}{\sin x}=\ln (\sin x)+C.
$$ 


*

*Integrating by parts we have


$$
\int \tanh^{-1}(x)\:dx=x \tanh^{-1}(x)-\int\frac{x}{1-x^2}=x \tanh^{-1}(x)+\frac12 \ln\left(1-x^2\right)+C.
$$ 


*

*We have


$$
\int\sec^2(x)\:dx=\int\frac1{\cos^2 x}dx=\tan x+C.
$$
