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

The integral from 0 to t of cos(x)sin(t-x) is (tsin(t))/2

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

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

\frac{t sin ( t )}{2}

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

\frac{1}{4} cos (t) + \frac{t sin ( t ) - cos ( t )}{4} + \frac{1}{4} t sin (t)

= \frac{1}{4} cos (t) + \frac{t sin ( t ) - cos ( t )}{4} + \frac{1}{4} t sin (t)

Simplify

= \frac{t sin ( t )}{2}

\int_{1}^{9} (\frac{2 ln ( x )}{x}) d x

\int_{1}^{3} \frac{1}{( x ^{2} + 3 ) ^{2}} d x

\int_{1}^{2} e^{- s t} t d t

\int_{3}^{7} \frac{1}{5 x + 1} d x

\int_{2}^{5} x^{\frac{1}{3}} (- 1 - 2 x) d x

What is the integral from 0 to t of cos(x)sin(t-x) ?
The integral from 0 to t of cos(x)sin(t-x) is (tsin(t))/2