CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

$$\displaystyle \int 2e^{x}\frac{\sqrt{1+sin2x}}{1+cos2x}dx=$$


A
exsecx+c
loader
B
ex tan x+c
loader
C
ex secx tan x+c
loader
D
ex cosec x+c
loader

Solution

The correct option is A $$ e^{x}secx+c$$
$$\int 2e^{x}\dfrac{\sqrt{1+sin2x}}{1+cos2x}dx=\int 2e^{x}\dfrac{sinx+cosx}{2cos^{2}x}dx$$
$$\int e^{x}\left ( tanx\ secx+secx \right )dx$$
it is in the form of
$$\int e^{x}\left ( f(x)+f^{1}(x) \right )dx$$
$$=e^{x}f(x)+c$$
so
$$\int e^{x}\left ( tanx\ secx+secx \right )dx=e^{x}secx+c$$
$$so,\int 2e^{x}\dfrac{\sqrt{1+sin2x}}{1+cos2x}dx=e^{x}secx+c$$
$$while\ x\ \epsilon ({-\pi}/{4}, {3\pi }/{4})$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image