Question

Find $B$. If $\int \frac{si{n}^{2}x-co{s}^{2}x}{si{n}^{2}xco{s}^{2}x}dx=\tan x+B+c$

Answer 1

$\int \frac{si{n}^{2}x-co{s}^{2}x}{si{n}^{2}xco{s}^{2}x}$dx

Separating the denominators, we get

$\int \frac{{\sin }^{2}x}{{\sin }^{2}x{\cos }^{2}x}-\frac{{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx$

$=\int {\sec }^{2}x-cose{c}^{2}xdx$

$=\tan x+\cot x+c$

Comapring with

$\tan x+B+c$

we get

$B=cotx$