What is the integral of (csc(θ))/(4csc(θ)-3cot(θ)+1) ?

The integral of (csc(θ))/(4csc(θ)-3cot(θ)+1) is sqrt(2/3)arctan(1/(sqrt(6))(7tan(θ/2)+1))+C

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integral of (csc(θ))/(4csc(θ)-3cot(θ)+1)

\int \frac{csc ( θ )}{4 csc ( θ ) - 3 cot ( θ ) + 1} d θ

\frac{2}{3} arctan (\frac{1}{6} (7 tan (\frac{θ}{2}) + 1)) + C

Solution steps

\int \frac{csc ( θ )}{4 csc ( θ ) - 3 cot ( θ ) + 1} d θ

Apply u-substitution

: \int \frac{2}{7 u ^{2} + 2 u + 1} d u

= \int \frac{2}{7 u ^{2} + 2 u + 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{1}{7 u ^{2} + 2 u + 1} d u

Complete the square

7 u^{2} + 2 u + 1 : 7 (u + \frac{1}{7})^{2} + \frac{6}{7}

= 2 \cdot \int \frac{1}{7 ( u + \frac{1}{7} ) ^{2} + \frac{6}{7}} d u

Apply u-substitution

: \int \frac{7}{49 v ^{2} + 6} d v

= 2 \cdot \int \frac{7}{49 v ^{2} + 6} d v

Take the constant out:

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

= 2 \cdot 7 \cdot \int \frac{1}{49 v ^{2} + 6} d v

Apply Integral Substitution

= 2 \cdot 7 \cdot \int \frac{1}{7 6 ( w ^{2} + 1 )} d w

Take the constant out:

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

= 2 \cdot 7 \cdot \frac{1}{7 6} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= 2 \cdot 7 \cdot \frac{1}{7 6} arctan (w)

Substitute back

= 2 \cdot 7 \cdot \frac{1}{7 6} arctan (\frac{7}{6} (tan (\frac{θ}{2}) + \frac{1}{7}))

Simplify

2 \cdot 7 \cdot \frac{1}{7 6} arctan (\frac{7}{6} (tan (\frac{θ}{2}) + \frac{1}{7})) : \frac{2}{3} arctan (\frac{1}{6} (7 tan (\frac{θ}{2}) + 1))

= \frac{2}{3} arctan (\frac{1}{6} (7 tan (\frac{θ}{2}) + 1))

Add a constant to the solution

= \frac{2}{3} arctan (\frac{1}{6} (7 tan (\frac{θ}{2}) + 1)) + C

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What is the integral of (csc(θ))/(4csc(θ)-3cot(θ)+1) ?
The integral of (csc(θ))/(4csc(θ)-3cot(θ)+1) is sqrt(2/3)arctan(1/(sqrt(6))(7tan(θ/2)+1))+C