What is the integral of sin^2(tco)s^4t ?

The integral of sin^2(tco)s^4t is (s^4)/(8c^2o^2)(2c^2o^2t^2-2cotsin(2cot)-cos(2cot))+C

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integral of sin^2(tco)s^4t

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

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

Solution steps

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

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 t) t d t

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 t

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

Simplify

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

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

Add a constant to the solution

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

\int f (3 3 x) d x

\frac{d}{d x} (4 x^{4} - 3 x^{3} - 30)

\int (1 - cos (2 x)) d x

\int \frac{x}{( x ^{2} + 6 x + 13 )} d x

x \to 2 lim (e^{2})

What is the integral of sin^2(tco)s^4t ?
The integral of sin^2(tco)s^4t is (s^4)/(8c^2o^2)(2c^2o^2t^2-2cotsin(2cot)-cos(2cot))+C