What is the integral of x^2(-cos(x))' ?

The integral of x^2(-cos(x))' is 1/3 x^3(-cos(x))'-1/3 x^3sin(x)-x^2cos(x)+2xsin(x)+2cos(x)+C

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integral of x^2(-cos(x))'

\int x^{2} (- cos (x))^{'} d x

\frac{1}{3} x^{3} (- cos (x))' - \frac{1}{3} x^{3} sin (x) - x^{2} cos (x) + 2 x sin (x) + 2 cos (x) + C

Solution steps

\int x^{2} (- cos (x))^{'} d x

Apply Integration By Parts

: \frac{1}{3} x^{3} (- cos (x))^{'} - \int \frac{1}{3} x^{3} cos (x) d x

= \frac{1}{3} x^{3} (- cos (x))^{^{'}} - \int \frac{1}{3} x^{3} cos (x) d x

\int \frac{1}{3} x^{3} cos (x) d x = \frac{1}{3} x^{3} sin (x) + x^{2} cos (x) - 2 x sin (x) - 2 cos (x)

= \frac{1}{3} x^{3} (- cos (x))^{^{'}} - (\frac{1}{3} x^{3} sin (x) + x^{2} cos (x) - 2 x sin (x) - 2 cos (x))

Simplify

= \frac{1}{3} x^{3} (- cos (x))^{^{'}} - \frac{1}{3} x^{3} sin (x) - x^{2} cos (x) + 2 x sin (x) + 2 cos (x)

Add a constant to the solution

= \frac{1}{3} x^{3} (- cos (x))^{^{'}} - \frac{1}{3} x^{3} sin (x) - x^{2} cos (x) + 2 x sin (x) + 2 cos (x) + C

\int t^{6} sin (2 t) d t

370 \cdot 2^{(\frac{1}{10} x)}

\int \frac{x}{( 5 x + 2 ) ^{3}} d x

\int_{0}^{x} arctan (t^{2}) d t

What is the integral of x^2(-cos(x))' ?
The integral of x^2(-cos(x))' is 1/3 x^3(-cos(x))'-1/3 x^3sin(x)-x^2cos(x)+2xsin(x)+2cos(x)+C