Question

$\int \frac{se{c}^{2}x}{(secx+tanx{)}^5}dx=$

• A. $\frac{1}{4(secx+tanx{)}^4}[\frac{1}{2}-\frac{1}{3(secx+tanx{)}^2}]+c$
• B. $\frac{1}{4(secx+tanx{)}^4}[\frac{1}{2}+\frac{1}{3(secx+tanx{)}^2}]+c$
• C. $-\frac{1}{4(secx+tanx{)}^4}[\frac{1}{2}+\frac{1}{3(secx-tanx{)}^2}]+c$
• D. None

Answer 1

The correct option is B $\frac{1}{4(secx+tanx{)}^4}[\frac{1}{2}+\frac{1}{3(secx+tanx{)}^2}]+c$

Put $t=secx+tanx\Rightarrow \frac{dt}{t}=secxdx$
$\int \frac{se{c}^{2}x}{(secx+tanx{)}^5}dx=\int \frac{\frac{1}{2}(t+1/t)}{t^5}\frac{dt}{t}=\frac{1}{2}\int (t^{-5}+t^{-7})dt=\frac{1}{2}[\frac{t^{-4}}{-4}+\frac{t^{-6}}{-6}]+c=-\frac{1}{2}[\frac{1}{4(secx+tanx{)}^4}+\frac{1}{6(secx+tanx{)}^6}]+c=\frac{-1}{4(secx+tanx{)}^4}[\frac{1}{2}+\frac{1}{3(secx+tanx{)}^2}]+c$