What is the integral of (csc(3x)+1)^2 ?

The integral of (csc(3x)+1)^2 is -1/3 cot(3x)+2/3 ln|tan((3x)/2)|+x+C

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integral of (csc(3x)+1)^2

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

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

Solution steps

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

Rewrite using trig identities

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

Expand

\frac{( 1 + sin ( 3 x ) ) ^{2}}{sin ^{2} ( 3 x )} : \frac{1}{sin ^{2} ( 3 x )} + \frac{2}{sin ( 3 x )} + 1

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

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

\int \frac{2}{sin ( 3 x )} d x = \frac{2}{3} ln tan (\frac{3 x}{2})

\int 1 d x = x

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

Add a constant to the solution

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

\int x d (tan (x)) d x

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\int \frac{1}{s ^{4} - 2 s ^{3} + s ^{2}} d s

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