What is the integral of 1/(sin(x)-2csc(x)) ?

The integral of 1/(sin(x)-2csc(x)) is arctan(sqrt(2)tan(x/2)+1)-arctan(sqrt(2)tan(x/2)-1)+C

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integral of 1/(sin(x)-2csc(x))

\int \frac{1}{sin ( x ) - 2 csc ( x )} d x

arctan (2 tan (\frac{x}{2}) + 1) - arctan (2 tan (\frac{x}{2}) - 1) + C

Solution steps

\int \frac{1}{sin ( x ) - 2 csc ( x )} d x

Apply u-substitution

: \int \frac{2 u}{- u ^{4} - 1} d u

= \int \frac{2 u}{- u ^{4} - 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{u}{- u ^{4} - 1} d u

Take the partial fraction of

\frac{u}{- u ^{4} - 1} : \frac{1}{2 2 ( u ^{2} + 2 u + 1 )} - \frac{1}{2 2 ( u ^{2} - 2 u + 1 )}

= 2 \cdot \int \frac{1}{2 2 ( u ^{2} + 2 u + 1 )} - \frac{1}{2 2 ( u ^{2} - 2 u + 1 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{1}{2 2 ( u ^{2} + 2 u + 1 )} d u - \int \frac{1}{2 2 ( u ^{2} - 2 u + 1 )} d u)

\int \frac{1}{2 2 ( u ^{2} + 2 u + 1 )} d u = \frac{1}{2} arctan (2 u + 1)

\int \frac{1}{2 2 ( u ^{2} - 2 u + 1 )} d u = \frac{1}{2} arctan (2 u - 1)

= 2 (\frac{1}{2} arctan (2 u + 1) - \frac{1}{2} arctan (2 u - 1))

Substitute back

u = tan (\frac{x}{2})

= 2 (\frac{1}{2} arctan (2 tan (\frac{x}{2}) + 1) - \frac{1}{2} arctan (2 tan (\frac{x}{2}) - 1))

Simplify

2 (\frac{1}{2} arctan (2 tan (\frac{x}{2}) + 1) - \frac{1}{2} arctan (2 tan (\frac{x}{2}) - 1)) : arctan (2 tan (\frac{x}{2}) + 1) - arctan (2 tan (\frac{x}{2}) - 1)

= arctan (2 tan (\frac{x}{2}) + 1) - arctan (2 tan (\frac{x}{2}) - 1)

Add a constant to the solution

= arctan (2 tan (\frac{x}{2}) + 1) - arctan (2 tan (\frac{x}{2}) - 1) + C

y = \frac{5 x}{x - 2}, (3, 15)

f (x) = \frac{x ^{2}}{x + 4}

\int (6 e^{x}) d x

n = 0 \sum \infty \frac{7}{2 ^{n}}

\frac{d}{d x} (e^{x} x^{6} + 12 e^{x} x^{5} + 30 e^{x} x^{4})

What is the integral of 1/(sin(x)-2csc(x)) ?
The integral of 1/(sin(x)-2csc(x)) is arctan(sqrt(2)tan(x/2)+1)-arctan(sqrt(2)tan(x/2)-1)+C