∫(2x+ sin2x/1+ cos2x) dx is equal to

Question

Tardigrade Admin · Accepted Answer

Let I=∫ (2 x/1+ cos 2 x) =∫ (2 x/1+ cos 2 x)+( sin 2 x/1+ cos 2 x) d x =∫ (2 x/2 cos 2 x)+∫ tan x d x =∫ x sec 2 x+∫ tan x d x =x tan x-∫ tan x d x+∫ tan x d x =x tan x+C

Q. $\int \frac{2 x + s i n 2 x}{1 + c o s 2 x} d x$ is equal to

$x + lo g ∣ tan x ∣ + C$

$x lo g ∣ tan x ∣ + C$

$x tan x + C$

$lo g ∣ cos x ∣ + C$

$lo g ∣ sin x ∣ + C$

Let $I = \int \frac{2 x}{1 + c o s 2 x}$
$= \int \frac{2 x}{1 + c o s 2 x} + \frac{s i n 2 x}{1 + c o s 2 x} d x$
$= \int \frac{2 x}{2 c o s ^{2} x} + \int tan x d x$
$= \int x sec^{2} x + \int tan x d x$
$= x tan x - \int tan x d x + \int tan x d x$
$= x tan x + C$

Q. ∫1+cos2x2x+sin2x​dx is equal to

x+log∣tanx∣+C

xlog∣tanx∣+C

xtanx+C

log∣cosx∣+C

log∣sinx∣+C

Let I=∫1+cos2x2x​ =∫1+cos2x2x​+1+cos2xsin2x​dx =∫2cos2x2x​+∫tanxdx =∫xsec2x+∫tanxdx =xtanx−∫tanxdx+∫tanxdx =xtanx+C

Q. $\int \frac{2 x + s i n 2 x}{1 + c o s 2 x} d x$ is equal to

$x + lo g ∣ tan x ∣ + C$

$x lo g ∣ tan x ∣ + C$

$x tan x + C$

$lo g ∣ cos x ∣ + C$

$lo g ∣ sin x ∣ + C$

Let $I = \int \frac{2 x}{1 + c o s 2 x}$
$= \int \frac{2 x}{1 + c o s 2 x} + \frac{s i n 2 x}{1 + c o s 2 x} d x$
$= \int \frac{2 x}{2 c o s ^{2} x} + \int tan x d x$
$= \int x sec^{2} x + \int tan x d x$
$= x tan x - \int tan x d x + \int tan x d x$
$= x tan x + C$