Question

$\int \frac{secx}{(secx+tanx{)}^2}dx=$

• A. $-\frac{1}{2(secx+tanx{)}^2}+c$
• B. $\frac{1}{(secx+tanx{)}^2}+c$
• C. $\frac{1}{2(secx-tanx{)}^2}+c$
• D. None

Answer 1

The correct option is A $-\frac{1}{2(secx+tanx{)}^2}+c$

Put $t=secx+tanx\Rightarrow \frac{dt}{t}=secxdx$
$\int \frac{secx}{(secx+tanx{)}^2}dx=\int \frac{1}{t^2}\frac{dt}{t}=\int t^{-3}dt=\frac{t^{-2}}{-2}+c=\frac{-1}{2(secx+tanx{)}^2}+c$