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

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

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integral of 2x^2cos(5x)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

= 2 \cdot \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

= 2 \cdot \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))

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

Substitute back

u = 5 x

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

Simplify

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

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

Add a constant to the solution

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

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x \to 0 + lim (2^{- \frac{1}{x}})

\int \frac{1}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}} d x

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