What is the integral from 0 to 1 of (1-x)^2sin(x) ?

The integral from 0 to 1 of (1-x)^2sin(x) is 2cos(1)-1

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integral from 0 to 1 of (1-x)^2sin(x)

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

2 cos (1) - 1

Decimal

0.08060 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int 2 cos (x) (1 - x) d x = 2 (sin (x) (1 - x) - cos (x))

= [- cos (x) (1 - x)^{2} - 2 (sin (x) (1 - x) - cos (x))]_{0}^{1}

Expand

- cos (x) (1 - x)^{2} - 2 (sin (x) (1 - x) - cos (x)) : - x^{2} cos (x) + 2 x cos (x) + 2 x sin (x) + cos (x) - 2 sin (x)

= [- x^{2} cos (x) + 2 x cos (x) + 2 x sin (x) + cos (x) - 2 sin (x)]_{0}^{1}

Compute the boundaries:

2 cos (1) - 1

= 2 cos (1) - 1

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

\int_{0}^{16} \frac{x}{49 + 2 x} d x

\int_{e}^{e^{2}} \frac{ln ^{2} ( x )}{x} d x

\int_{0}^{1} 8 x^{2} ln (x) d x

\int_{3}^{5} x (2 - x) d x

What is the integral from 0 to 1 of (1-x)^2sin(x) ?
The integral from 0 to 1 of (1-x)^2sin(x) is 2cos(1)-1