The integral ∫ √ 1 + 2 cot x (cosec x + cot x) dx ( 0

Question

The integral ∫ √ 1 + 2 cot x (cosec x + cot x) dx ( 0 < x < (π/2) ) is equal to: (where C is a constant of integration) (A) 4 log( sin (x/2)) + C  (B) 2 log( sin (x/2)) + C  (C) 2 log( cos(x/2)) + C  (D) 4 log( cos(x/2)) + C

Tardigrade Admin · Accepted Answer

∫(√+2cot x cos ecx + cos ec2x + cot x)dx ∫ cos | x + cot x | dx ∫(cos ec + cot x)dx ∫ cosec dx 2log(log(x2 ) + c)

Q. The integral $\int 1 + 2 cot x (cosec x + cot x) d x (0 < x < \frac{π}{2})$ is equal to :
(where $C$ is a constant of integration)

$4 lo g (sin \frac{x}{2}) + C$

$2 lo g (sin \frac{x}{2}) + C$

$2 lo g (cos \frac{x}{2}) + C$

$4 lo g (cos \frac{x}{2}) + C$

$\int (+ 2 co t x cosec x + cose c^{2} x + co t x) d x$
$\int cos ∣ x + co t x ∣ d x$
$\int (cosec + co t x) d x$
$\int cosec d x$
$2 l o g (l o g (x_{2}) + c)$

Q. The integral ∫1+2cotx(cosecx+cotx)​dx(0<x<2π​) is equal to : (where C is a constant of integration)

4log(sin2x​)+C

2log(sin2x​)+C

2log(cos2x​)+C

4log(cos2x​)+C