$$\int \frac{1+\tan x \tan (x+a)}{\tan x \tan (x+a)} d x=$$

Assertion (A) If $$I_n=\int \cot ^n x d x$$, then $$I_6+I_4=\frac{-\cot ^5 x}{5}$$

Reason (R) $$\int \cot ^n x d x=\frac{-\cot ^{n-1} x}{n} -\int \cot ^{n-2} x d x$$

If $$I_n=\int \tan ^n x d x$$, and $$I_0+I_1+2 I_2+2 I_3+2 I_4 +I_5+I_6=\sum_\limits{k=1}^n \frac{\tan ^k x}{k}$$, then $$n=$$

$$\int \frac{e^{\cot x}}{\sin ^2 x}(2 \log \operatorname{cosec} x+\sin 2 x) d x=$$