What is the integral of sin^2(xco)s^4x ?

The integral of sin^2(xco)s^4x is (s^4)/(8c^2o^2)(2c^2o^2x^2-2coxsin(2cox)-cos(2cox))+C

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integral of sin^2(xco)s^4x

\int sin^{2} (x co) s^{4} x d x

\frac{s ^{4}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x)) + C

Solution steps

\int sin^{2} (x co) s^{4} x d x

Take the constant out:

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

= s^{4} \cdot \int sin^{2} (co x) x d x

Apply u-substitution

: \int \frac{u sin ^{2} ( u )}{c ^{2} o ^{2}} d u

= s^{4} \cdot \int \frac{u sin ^{2} ( u )}{c ^{2} o ^{2}} d u

Take the constant out:

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

= s^{4} \frac{1}{c ^{2} o ^{2}} \cdot \int u sin^{2} (u) d u

Apply Integration By Parts

: \frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u

= s^{4} \frac{1}{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))

= s^{4} \frac{1}{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 = co x

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

Simplify

s^{4} \frac{1}{c ^{2} o ^{2}} (\frac{1}{2} co x (co x - \frac{1}{2} sin (2 co x)) - \frac{1}{2} (\frac{( co x ) ^{2}}{2} + \frac{1}{4} cos (2 co x))) : \frac{s ^{4}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x))

= \frac{s ^{4}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x))

Add a constant to the solution

= \frac{s ^{4}}{8 c ^{2} o ^{2}} (2 c^{2} o^{2} x^{2} - 2 co x sin (2 co x) - cos (2 co x)) + C

x \to 0 + lim (sin (x) ln (3 x))

x^{2} - x + 4

\frac{d x}{d y} - x y = x^{2} y^{3}

x \to \infty lim (x^{- e})

f (x) = 2 + \frac{1}{x}, (2, 2.5)

What is the integral of sin^2(xco)s^4x ?
The integral of sin^2(xco)s^4x is (s^4)/(8c^2o^2)(2c^2o^2x^2-2coxsin(2cox)-cos(2cox))+C