How do you find the indefinite integral of $\int csc 2 x$ $int csc2x$ ?

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Answer 1 · 2017-05-07T08:38:43

$\int csc x 2 x d x = - \frac{1}{2} ln | (csc 2 x + cot 2 x) | + C$ $intcscx2xdx=-1/2ln| (csc2x+cot2x)|+C$

start by multiplying by $\frac{csc 2 x + cot 2 x}{csc 2 x + cot 2 x}$ $(csc2x+cot2x)/(csc2x+cot2x)$

$\int csc x 2 x d x = \int csc 2 x \times \frac{csc 2 x + cot 2 x}{csc 2 x + cot 2 x} d x$ $intcscx2xdx=intcsc 2x xx(csc2x+cot2x)/(csc2x+cot2x)dx$

$= \int [\frac{{csc}^{2} 2 x + csc 2 x cot 2 x}{csc 2 x + cot 2 x}] d x$ $=int[(csc^2 2x+csc2xcot2x)/(csc2x+cot2x)]dx$

now

$\frac{d}{d x} (csc 2 x) = - 2 cot 2 x csc 2 x$ $d/(dx)(csc2x)=-2cot2xcsc2x$

and

$\frac{d}{d x} (cot 2 x) = - 2 {csc}^{2} 2 x$ $d/(dx)(cot2x)=-2csc^2 2x$

also

$int((f'(x))/f(x))dx=ln|f(x)|+C$

using these results we notice that differentiating the denominator:

$d/(dx)(csc2x+cot2x)=-2cot2xcsc2x-2csc^2 2x$

$=-2(csc^2 2x+csc2x+cot2x)=-2 xx" numerator"$

$:. intcscx2xdx=-1/2ln| (csc2x+cot2x)|+C$

How do you find the indefinite integral of int csc2x∫csc2xint csc2x?