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

The integral of csc^2(3x)cos(3x) is -1/3 csc(3x)+C

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integral of csc^2(3x)cos(3x)

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

- \frac{1}{3} csc (3 x) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{3} \cdot \int csc^{2} (u) cos (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \frac{1}{3} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{3} (- \frac{1}{v})

Substitute back

= \frac{1}{3} (- \frac{1}{sin ( 3 x )})

Simplify

\frac{1}{3} (- \frac{1}{sin ( 3 x )}) : - \frac{1}{3} csc (3 x)

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

Add a constant to the solution

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

x \to 0 lim (x - 2)

y^{^{''}} - 6 y^{^{'}} - 2 y + 1 = cos (2 x)

(x^{2} + 4) \frac{d y}{d x} + 3 x y = x

\int (x - 1)^{4} d x

\int 2 x + 7 d x

What is the integral of csc^2(3x)cos(3x) ?
The integral of csc^2(3x)cos(3x) is -1/3 csc(3x)+C