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

The integral of 1/(sin(x)(2+cos(x)-2sin(x))) is

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integral of 1/(sin(x)(2+cos(x)-2sin(x)))

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

\frac{1}{2} ln tan^{2} (\frac{x}{2}) - 4 tan (\frac{x}{2}) + 3 + \frac{1}{3} ln tan (\frac{x}{2}) - \frac{3}{2} ln tan (\frac{x}{2}) - 1 + \frac{7}{6} ln tan (\frac{x}{2}) - 3 + C

Solution steps

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

Rewrite using trig identities

= \int \frac{csc ( x )}{2 + cos ( x ) - 2 sin ( x )} d x

Apply u-substitution

: \int \frac{u ^{2} + 1}{u ( u ^{2} - 4 u + 3 )} d u

= \int \frac{u ^{2} + 1}{u ( u ^{2} - 4 u + 3 )} d u

Expand

\frac{u ^{2} + 1}{u ( u ^{2} - 4 u + 3 )} : \frac{u ^{2}}{u ( u ^{2} - 4 u + 3 )} + \frac{1}{u ( u ^{2} - 4 u + 3 )}

Apply the Sum Rule:

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

= \int \frac{u ^{2}}{u ( u ^{2} - 4 u + 3 )} d u + \int \frac{1}{u ( u ^{2} - 4 u + 3 )} d u

\int \frac{u ^{2}}{u ( u ^{2} - 4 u + 3 )} d u = \frac{1}{2} ln u^{2} - 4 u + 3 - ln ∣ u - 1 ∣ + ln ∣ u - 3 ∣

\int \frac{1}{u ( u ^{2} - 4 u + 3 )} d u = \frac{1}{3} ln ∣ u ∣ - \frac{1}{2} ln ∣ u - 1 ∣ + \frac{1}{6} ln ∣ u - 3 ∣

= \frac{1}{2} ln u^{2} - 4 u + 3 - ln ∣ u - 1 ∣ + ln ∣ u - 3 ∣ + \frac{1}{3} ln ∣ u ∣ - \frac{1}{2} ln ∣ u - 1 ∣ + \frac{1}{6} ln ∣ u - 3 ∣

Substitute back

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

= \frac{1}{2} ln tan^{2} (\frac{x}{2}) - 4 tan (\frac{x}{2}) + 3 - ln tan (\frac{x}{2}) - 1 + ln tan (\frac{x}{2}) - 3 + \frac{1}{3} ln tan (\frac{x}{2}) - \frac{1}{2} ln tan (\frac{x}{2}) - 1 + \frac{1}{6} ln tan (\frac{x}{2}) - 3

Simplify

\frac{1}{2} ln tan^{2} (\frac{x}{2}) - 4 tan (\frac{x}{2}) + 3 - ln tan (\frac{x}{2}) - 1 + ln tan (\frac{x}{2}) - 3 + \frac{1}{3} ln tan (\frac{x}{2}) - \frac{1}{2} ln tan (\frac{x}{2}) - 1 + \frac{1}{6} ln tan (\frac{x}{2}) - 3 : \frac{1}{2} ln tan^{2} (\frac{x}{2}) - 4 tan (\frac{x}{2}) + 3 + \frac{1}{3} ln tan (\frac{x}{2}) - \frac{3}{2} ln tan (\frac{x}{2}) - 1 + \frac{7}{6} ln tan (\frac{x}{2}) - 3

= \frac{1}{2} ln tan^{2} (\frac{x}{2}) - 4 tan (\frac{x}{2}) + 3 + \frac{1}{3} ln tan (\frac{x}{2}) - \frac{3}{2} ln tan (\frac{x}{2}) - 1 + \frac{7}{6} ln tan (\frac{x}{2}) - 3

Add a constant to the solution

= \frac{1}{2} ln tan^{2} (\frac{x}{2}) - 4 tan (\frac{x}{2}) + 3 + \frac{1}{3} ln tan (\frac{x}{2}) - \frac{3}{2} ln tan (\frac{x}{2}) - 1 + \frac{7}{6} ln tan (\frac{x}{2}) - 3 + C

\int \frac{x ^{2}}{2 + 2 x ^{2}} d x

\int \frac{x}{x ^{2} - 6 x + 9} d x

\int 2 x^{2} (x^{3} + 8)^{5} d x

\int \frac{x}{( x ^{2} - 2 x + 2 ) ^{2}} d x

\int \frac{ln ( x )}{ln ( x )} d x

What is the integral of 1/(sin(x)(2+cos(x)-2sin(x))) ?
The integral of 1/(sin(x)(2+cos(x)-2sin(x))) is