What is the integral of x^2cos(5x) ?

The integral of x^2cos(5x) is 1/125 (25x^2sin(5x)-2(-5xcos(5x)+sin(5x)))+C

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integral of x^2cos(5x)

\int x^{2} cos (5 x) d x

\frac{1}{125} (25 x^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x))) + C

Solution steps

\int x^{2} cos (5 x) d x

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{125} d u

= \int \frac{u ^{2} cos ( u )}{125} d u

Take the constant out:

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

= \frac{1}{125} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

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

= \frac{1}{125} (u^{2} sin (u) - \int 2 u sin (u) d u)

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= \frac{1}{125} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 5 x

= \frac{1}{125} ((5 x)^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x)))

Simplify

\frac{1}{125} ((5 x)^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x))) : \frac{1}{125} (25 x^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x)))

= \frac{1}{125} (25 x^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x)))

Add a constant to the solution

= \frac{1}{125} (25 x^{2} sin (5 x) - 2 (- 5 x cos (5 x) + sin (5 x))) + C

f (x) = 2 sin (x), (\frac{π}{6}, 1)

\int 2 x e^{x^{3}} d x

x y y^{^{'}} = y - 1

e^{- \frac{7}{x ^{7}}}

y = \frac{1}{cos ^{2} ( x )}

What is the integral of x^2cos(5x) ?
The integral of x^2cos(5x) is 1/125 (25x^2sin(5x)-2(-5xcos(5x)+sin(5x)))+C