Question

Solve the integral:

$\int \frac{\tan x}{\sec x+\tan x}dx$=

• A. $x+\sec x+\tan x+c$
• B. $2x-\tan x+\sec x+c$
• C. $x-\tan x+\sec x+c$
• D. None of the above

Answer 1

The correct option is C $x-\tan x+\sec x+c$

$\int \left(\frac{tanx}{secx+tanx}\right)dx\int \left(\frac{secx+tanx-secx}{secx+tanx}\right)dx=\int 1-\left(\frac{secx}{secx+tanx}\right)dxsecx+tanx=t(tanx+se{c}^{2}x)dx=dt\therefore secxdx=\left(\frac{dt}{secx+tanx}\right)dxI=\int 1dx-\int \left(\frac{dt}{(secx+tanx{)}^2}\right)=x-\int \left(\frac{1}{t^2}\right)dt=x+\left(\frac{1}{t}\right)+c=x+\left(\frac{1}{secx+tanx}\right)+c=x+secx-tanx+c$