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

The integral from 0 to pi/4 of x^2sin(4x) is (pi^2-2)/(64)-1/32

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{π ^{2} - 2}{64} - \frac{1}{32}

Decimal Notation

0.09171 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{4} (- x^{2} cos (4 x) - 4 \cdot \int - \frac{1}{2} x cos (4 x) d x)]_{0}^{\frac{π}{4}}

= [\frac{1}{4} (- x^{2} cos (4 x) - 4 \cdot \int - \frac{1}{2} x cos (4 x) d x)]_{0}^{\frac{π}{4}}

\int - \frac{1}{2} x cos (4 x) d x = - \frac{1}{32} (4 x sin (4 x) + cos (4 x))

= [\frac{1}{4} (- x^{2} cos (4 x) - 4 (- \frac{1}{32} (4 x sin (4 x) + cos (4 x))))]_{0}^{\frac{π}{4}}

Simplify

[\frac{1}{4} (- x^{2} cos (4 x) - 4 (- \frac{1}{32} (4 x sin (4 x) + cos (4 x))))]_{0}^{\frac{π}{4}} : [\frac{1}{4} (- x^{2} cos (4 x) + \frac{1}{8} (4 x sin (4 x) + cos (4 x)))]_{0}^{\frac{π}{4}}

= [\frac{1}{4} (- x^{2} cos (4 x) + \frac{1}{8} (4 x sin (4 x) + cos (4 x)))]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{π ^{2} - 2}{64} - \frac{1}{32}

= \frac{π ^{2} - 2}{64} - \frac{1}{32}

\int_{0}^{4} 6 d x

\int_{2}^{s i n (x)} (t^{2} - 1) d t

\int_{0}^{1} 2 x arctan (x) d x

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

\int_{x^{2}}^{x^{4}} u d u

What is the integral from 0 to pi/4 of x^2sin(4x) ?
The integral from 0 to pi/4 of x^2sin(4x) is (pi^2-2)/(64)-1/32