What is the integral of sin^2(3xco)s^23x ?

The integral of sin^2(3xco)s^23x is (s^2)/(24c^2o^2)(18c^2o^2x^2-6coxsin(6cox)-cos(6cox))+C

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integral of sin^2(3xco)s^23x

\int sin^{2} (3 x co) s^{2} 3 x d x

\frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x)) + C

Solution steps

\int sin^{2} (3 x co) s^{2} \cdot 3 x d x

Take the constant out:

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

= 3 s^{2} \cdot \int sin^{2} (3 co x) x d x

Apply u-substitution

= 3 s^{2} \cdot \int \frac{u sin ^{2} ( u )}{9 c ^{2} o ^{2}} d u

Take the constant out:

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

= 3 s^{2} \frac{1}{9 c ^{2} o ^{2}} \cdot \int u sin^{2} (u) d u

Apply Integration By Parts

= 3 s^{2} \frac{1}{9 c ^{2} o ^{2}} (\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u)

\int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u = \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))

= 3 s^{2} \frac{1}{9 c ^{2} o ^{2}} (\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u)))

Substitute back

u = 3 co x

= 3 s^{2} \frac{1}{9 c ^{2} o ^{2}} (\frac{1}{2} \cdot 3 co x (3 co x - \frac{1}{2} sin (2 \cdot 3 co x)) - \frac{1}{2} (\frac{( 3 co x ) ^{2}}{2} + \frac{1}{4} cos (2 \cdot 3 co x)))

Simplify

3 s^{2} \frac{1}{9 c ^{2} o ^{2}} (\frac{1}{2} \cdot 3 co x (3 co x - \frac{1}{2} sin (2 \cdot 3 co x)) - \frac{1}{2} (\frac{( 3 co x ) ^{2}}{2} + \frac{1}{4} cos (2 \cdot 3 co x))) : \frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x))

= \frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x))

Add a constant to the solution

= \frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x)) + C

What is the integral of sin^2(3xco)s^23x ?
The integral of sin^2(3xco)s^23x is (s^2)/(24c^2o^2)(18c^2o^2x^2-6coxsin(6cox)-cos(6cox))+C