What is the integral of (t-x)^3sin(x) ?

The integral of (t-x)^3sin(x) is 1/4 sin(x)(-x+t)^4+C

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integral of (t-x)^3sin(x)

integral

(t - x)^{3} sin (x)

\frac{1}{4} sin (x) (- x + t)^{4} + C

Solution steps

\int (t - x)^{3} sin (x) d t

Take the constant out:

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

= sin (x) \cdot \int (- x + t)^{3} d t

Apply u-substitution

: \int u^{3} d u

= sin (x) \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= sin (x) \frac{u ^{4}}{4}

Substitute back

u = - x + t

= sin (x) \frac{( - x + t ) ^{4}}{4}

Simplify

sin (x) \frac{( - x + t ) ^{4}}{4} : \frac{1}{4} sin (x) (- x + t)^{4}

= \frac{1}{4} sin (x) (- x + t)^{4}

Add a constant to the solution

= \frac{1}{4} sin (x) (- x + t)^{4} + C

\frac{1}{x ^{y}}

((\frac{8}{x}) + 2 x)^{2}

\frac{x}{( a ^{2} - x ^{2} )}

3 e^{2} x

\frac{1}{( a ^{2} + b ^{2} )}

What is the integral of (t-x)^3sin(x) ?
The integral of (t-x)^3sin(x) is 1/4 sin(x)(-x+t)^4+C