What is the integral from 0 to pi/2 of x^2cos(4x) ?

The integral from 0 to pi/2 of x^2cos(4x) is pi/(16)

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integral from 0 to pi/2 of x^2cos(4x)

\int_{0}^{\frac{π}{2}} x^{2} cos (4 x) d x

\frac{π}{16}

Decimal Notation

0.19634 \dots

Solution steps

\int_{0}^{\frac{π}{2}} x^{2} cos (4 x) d x

Apply u-substitution

: \int_{0}^{2 π} \frac{u ^{2} cos ( u )}{64} d u

= \int_{0}^{2 π} \frac{u ^{2} cos ( u )}{64} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int_{0}^{2 π} u^{2} cos (u) d u

Apply Integration By Parts

: [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{2 π}

= \frac{1}{64} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{2 π}

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

= \frac{1}{64} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{2 π}

Compute the boundaries:

4 π

= \frac{1}{64} \cdot 4 π

Simplify

= \frac{π}{16}

\int_{0}^{5} 3 x + 2 d x

\int_{2}^{3} 3 t^{2} + 1 d t

\int_{0}^{12} \frac{1}{1 + x ^{2}} d x

\int_{3}^{\infty} \frac{3}{x ^{6}} d x

\int_{3}^{5} 3 + x^{2} d x

What is the integral from 0 to pi/2 of x^2cos(4x) ?
The integral from 0 to pi/2 of x^2cos(4x) is pi/(16)