#int(1+cos(2x))/(sin(2x))dx=?

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Answer 1 · 2018-04-14T15:07:22

The answer is $option (B)$ $"option (B)"$

We need

$\int (cot x) = ln | sin x | + C$ $int(cotx)=ln|sinx|+C$

$\int csc x = ln (| csc x + cot x |) + C$ $intcscx=ln(|cscx+cotx|)+C$

Let $u = 2 x$ $u=2x$ , $\Rightarrow$ $=>$ , $d u = 2 d x$ $du=2dx$

The integral is

$I = \int \frac{(1 + cos 2 x) d x}{sin 2 x} = \frac{1}{2} \int \frac{(1 + cos u) d u}{sin u}$ $I=int((1+cos2x)dx)/(sin2x)=1/2int((1+cosu)du)/(sinu)$

$= \frac{1}{2} \int (cot u + csc u) d u$ $=1/2int(cotu+cscu)du$

$= \frac{1}{2} (ln (sin u)) - \frac{1}{2} ln (csc u + cot u)$ $=1/2(ln(sinu))-1/2ln(cscu+cotu)$

$= \frac{1}{2} ln (sin u) - \frac{1}{2} ln (\frac{1 + cos u}{sin u})$ $=1/2ln(sinu)-1/2ln((1+cosu)/sinu)$

$= \frac{1}{2} ln (sin 2 x) - \frac{1}{2} ln (1 + cos 2 x) - \frac{1}{2} ln (sin 2 x) + C$ $=1/2ln(sin2x)-1/2ln(1+cos2x)-1/2ln(sin2x)+C$

$= - \frac{1}{2} ln (1 + 2 {cos}^{2} (2 x) - 1)$ $=-1/2ln(1+2cos^2 (2x)-1)$

$= - \frac{1}{2} \cdot 2 ln cos (2 x) + C$ $=-1/2*2lncos(2x)+C$

$= - ln (| cos 2 x |) + C$ $=-ln(|cos2x|)+C$

The answer is $option (B)$ $"option (B)"$