Question

$\int_{}^{}{\frac{\sec^{2}x\tan x}{\sec^{2}x + \tan^{2}x}dx}$ =

• A. log(sec² x + tan² x) + c
• B. $\frac{(\sec^{2}x + \tan^{2}x)^{2}}{2}$ + c
• C. $\frac{1}{2}$log (sec² x + tan²x) + c
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

$\int \frac { \sec ^ { 2 } x \tan x } { \sec ^ { 2 } x + \tan ^ { 2 } x } d x$ t = sec² x + tan²x

= $\int \frac { \mathrm { dt } / 4 } { \mathrm { t } }$ $\frac{dt}{dx}$=2secx.secx tan x + 2 tan x.sec² x

=$\frac{1}{4}$ log (sec² x + tan²x) + c = 4 tan x sec²x

$\frac{dt}{4}$ = tan x sec²x dx