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

The integral of 1/((2+cos(x))sin(x)) is 1/3 ln|tan(x/2)|+1/3 ln|tan^2(x/2)+3|+C

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

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

\frac{1}{3} ln tan (\frac{x}{2}) + \frac{1}{3} ln tan^{2} (\frac{x}{2}) + 3 + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

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

Substitute back

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

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

Add similar elements:

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

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

Add a constant to the solution

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

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1

\int \frac{2}{( x + 1 ) ( x + 3 )} d x

What is the integral of 1/((2+cos(x))sin(x)) ?
The integral of 1/((2+cos(x))sin(x)) is 1/3 ln|tan(x/2)|+1/3 ln|tan^2(x/2)+3|+C