Question

∫ eˣ(1+sin x/1+cos x) dx=......+c ,Select correct option from the given options.
(a) eˣ cot x
(b) eˣ cot x/2
(c) eˣ tan x/2
(d) eˣ/². tan x/2

Answer 1

HELLO DEAR,



∫e^x(1 + sinx)/(1 + cosx).dx

= ∫e^x*[1/(1 + cosx) + sinx/(1 + cosx)],dx

= ∫e^x*[1/2cos²(x/2) + 2sin(x/2)cos(x/2)/2cos²(x/2)],dx

= ∫e^x[{sec²(x/2)}/2 + tan(x/2)]

= ∫e^x[tan(x/2) + {sec²(x/2)}/2]

so, we know if integral is in the form of e^x{f(x) + f'(x)] so, your integral is e^xf(x) + c.

therefore, e^xtan(x/2) + c.


hence, option (c) is correct,


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

we have to find the value of $\int{e^x\{\frac{1+sinx}{1+cosx}\}}\,dx$

we know, $sin2\theta=\frac{2tan\theta}{1+tan^2\theta}$
and $cos2\theta=\frac{1-tan^2\theta}{1+tan^2\theta}$

so, $sinx=\frac{2tanx/2}{1+tan^2x/2}$

$\implies1+sinx=\frac{1+tan^2x/2+2tanx/2}{1+tan^2x/2}$

$\implies1+sinx=\frac{(1+tanx/2)^2}{1+tan^2x/2}$

similarly, $1+cosx=\frac{2}{1+tan^2x/2}$

hence,$\frac{1+sinx}{1+cosx}=\frac{(1+tanx/2)^2}{2}=\frac{1}{2}[1+tan^2x/2+2tanx/2]$

$=\frac{1}{2}[sec^2x/2+2tanx/2]$

now, $\int{e^x\{\frac{1}{2}[sec^2x/2+2tanx/2]\}}\,dx$

if we assume 2tanx/2 = f(x)
then, sec²x/2 = f'(x)
e.g., $\int{e^x\{f'(x)+f(x)\}}\,dx=e^f(x)+C$

hence, $\int{e^x\{\frac{1}{2}[sec^2x/2+2tanx/2]\}}\,dx$

= $\frac{1}{2}\int{e^x\{(2tanx/2)+(sec^2x/2)\}}\,dx$

= $\frac{1}{2}e^x(2tanx/2)+C$

=$e^xtanx/2+C$

hence, option (c) is correct.