What is the integral of 5x^4cos(x) ?

The integral of 5x^4cos(x) is 5(x^4sin(x)-4(-x^3cos(x)+3(x^2sin(x)-2(-xcos(x)+sin(x)))))+C

integral of 5x^4cos(x)

\int 5 x^{4} cos (x) d x

5 (x^{4} sin (x) - 4 (- x^{3} cos (x) + 3 (x^{2} sin (x) - 2 (- x cos (x) + sin (x))))) + C

Solution steps

\int 5 x^{4} cos (x) d x

Take the constant out:

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

= 5 \cdot \int x^{4} cos (x) d x

Apply Integration By Parts

= 5 (x^{4} sin (x) - \int 4 x^{3} sin (x) d x)

\int 4 x^{3} sin (x) d x = 4 (- x^{3} cos (x) + 3 (x^{2} sin (x) - 2 (- x cos (x) + sin (x))))

= 5 (x^{4} sin (x) - 4 (- x^{3} cos (x) + 3 (x^{2} sin (x) - 2 (- x cos (x) + sin (x)))))

Add a constant to the solution

= 5 (x^{4} sin (x) - 4 (- x^{3} cos (x) + 3 (x^{2} sin (x) - 2 (- x cos (x) + sin (x))))) + C

\int (x^{\frac{5}{9}} - 7 x^{\frac{7}{6}}) d x

\int x^{7} sec^{2} (x^{8}) d x

\int \frac{10 ( x - 9 )}{( x + 5 ) ( x - 2 )} d x

\int \frac{1}{x ^{2}} + \frac{1}{x ^{3}} d x

\int \frac{x ^{2} - 5 x + 2}{x ^{3} - x ^{2} - x + 1} d x

What is the integral of 5x^4cos(x) ?
The integral of 5x^4cos(x) is 5(x^4sin(x)-4(-x^3cos(x)+3(x^2sin(x)-2(-xcos(x)+sin(x)))))+C