What is the integral from 0 to t of (t-x)sin(x) ?

The integral from 0 to t of (t-x)sin(x) is t-sin(t)

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integral from 0 to t of (t-x)sin(x)

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

t - sin (t)

Solution steps

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

Expand

(t - x) sin (x) : t sin (x) - x sin (x)

= \int_{0}^{t} t sin (x) - x sin (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{t} t sin (x) d x - \int_{0}^{t} x sin (x) d x

\int_{0}^{t} t sin (x) d x = t (- cos (t) + 1)

\int_{0}^{t} x sin (x) d x = sin (t) - t cos (t)

= t (- cos (t) + 1) - (sin (t) - t cos (t))

Simplify

t (- cos (t) + 1) - (sin (t) - t cos (t)) : t - sin (t)

= t - sin (t)

y^{^{'}} + \frac{2}{x} y = 6 \frac{y ^{3}}{x ^{2}}

f (x) = 5 x (1 + x^{2})

3 e^{x} x

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

\int \frac{1}{x ^{2} - 8 x + 25} d x

What is the integral from 0 to t of (t-x)sin(x) ?
The integral from 0 to t of (t-x)sin(x) is t-sin(t)