What is the integral of (3csc(3x)+2cot(3x))^2 ?

The integral of (3csc(3x)+2cot(3x))^2 is -4x-13/3 cot(3x)-4csc(3x)+C

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integral of (3csc(3x)+2cot(3x))^2

\int (3 csc (3 x) + 2 cot (3 x))^{2} d x

- 4 x - \frac{13}{3} cot (3 x) - 4 csc (3 x) + C

Solution steps

\int (3 csc (3 x) + 2 cot (3 x))^{2} d x

Rewrite using trig identities

= \int \frac{( 3 + 2 cos ( 3 x ) ) ^{2}}{sin ^{2} ( 3 x )} d x

Expand

\frac{( 3 + 2 cos ( 3 x ) ) ^{2}}{sin ^{2} ( 3 x )} : \frac{9}{sin ^{2} ( 3 x )} + \frac{12 cos ( 3 x )}{sin ^{2} ( 3 x )} + \frac{4 cos ^{2} ( 3 x )}{sin ^{2} ( 3 x )}

= \int \frac{9}{sin ^{2} ( 3 x )} + \frac{12 cos ( 3 x )}{sin ^{2} ( 3 x )} + \frac{4 cos ^{2} ( 3 x )}{sin ^{2} ( 3 x )} d x

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{9}{sin ^{2} ( 3 x )} d x + \int \frac{12 cos ( 3 x )}{sin ^{2} ( 3 x )} d x + \int \frac{4 cos ^{2} ( 3 x )}{sin ^{2} ( 3 x )} d x

\int \frac{9}{sin ^{2} ( 3 x )} d x = - 3 cot (3 x)

\int \frac{12 cos ( 3 x )}{sin ^{2} ( 3 x )} d x = - 4 csc (3 x)

\int \frac{4 cos ^{2} ( 3 x )}{sin ^{2} ( 3 x )} d x = 4 (- x - \frac{1}{3} cot (3 x))

= - 3 cot (3 x) - 4 csc (3 x) + 4 (- x - \frac{1}{3} cot (3 x))

Simplify

- 3 cot (3 x) - 4 csc (3 x) + 4 (- x - \frac{1}{3} cot (3 x)) : - 4 x - \frac{13}{3} cot (3 x) - 4 csc (3 x)

= - 4 x - \frac{13}{3} cot (3 x) - 4 csc (3 x)

Add a constant to the solution

= - 4 x - \frac{13}{3} cot (3 x) - 4 csc (3 x) + C

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

\int \frac{y}{( y ^{2} + 1 )} d y

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

\int \frac{3 x}{x} d x

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

What is the integral of (3csc(3x)+2cot(3x))^2 ?
The integral of (3csc(3x)+2cot(3x))^2 is -4x-13/3 cot(3x)-4csc(3x)+C