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

The integral of 1/((sin(2x)+2sin(x))) is 1/4 ((tan^2(x/2))/2+ln|tan(x/2)|)+C

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

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

\frac{1}{4} (\frac{tan ^{2} ( \frac{x}{2} )}{2} + ln tan (\frac{x}{2})) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{u ^{2} + 1}{u} : u + \frac{1}{u}

= \frac{1}{4} \cdot \int u + \frac{1}{u} 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

= \frac{1}{4} (\int u d u + \int \frac{1}{u} d u)

\int u d u = \frac{u ^{2}}{2}

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{4} (\frac{u ^{2}}{2} + ln ∣ u ∣)

Substitute back

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

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

Add a constant to the solution

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

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What is the integral of 1/((sin(2x)+2sin(x))) ?
The integral of 1/((sin(2x)+2sin(x))) is 1/4 ((tan^2(x/2))/2+ln|tan(x/2)|)+C