What is the integral of 3x^2cos(7x) ?

The integral of 3x^2cos(7x) is 3/343 (49x^2sin(7x)-2(-7xcos(7x)+sin(7x)))+C

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integral of 3x^2cos(7x)

\int 3 x^{2} cos (7 x) d x

\frac{3}{343} (49 x^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x))) + C

Solution steps

\int 3 x^{2} cos (7 x) d x

Take the constant out:

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

= 3 \cdot \int x^{2} cos (7 x) d x

Apply u-substitution

: \int \frac{u ^{2} cos ( u )}{343} d u

= 3 \cdot \int \frac{u ^{2} cos ( u )}{343} d u

Take the constant out:

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

= 3 \cdot \frac{1}{343} \cdot \int u^{2} cos (u) d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{343} (u^{2} sin (u) - \int 2 u sin (u) d u)

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

= 3 \cdot \frac{1}{343} (u^{2} sin (u) - 2 (- u cos (u) + sin (u)))

Substitute back

u = 7 x

= 3 \cdot \frac{1}{343} ((7 x)^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x)))

Simplify

3 \cdot \frac{1}{343} ((7 x)^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x))) : \frac{3}{343} (49 x^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x)))

= \frac{3}{343} (49 x^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x)))

Add a constant to the solution

= \frac{3}{343} (49 x^{2} sin (7 x) - 2 (- 7 x cos (7 x) + sin (7 x))) + C

\int 9 x d x

\int_{0}^{20} 2^{\frac{t}{10}} d t

\frac{\partial}{\partial x} (- 2 + e^{- x} cos (y))

\frac{10}{5 x}

\frac{( x - 1 )}{x + 1}

What is the integral of 3x^2cos(7x) ?
The integral of 3x^2cos(7x) is 3/343 (49x^2sin(7x)-2(-7xcos(7x)+sin(7x)))+C