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

The integral from 0 to x of tcos(t) is xsin(x)+cos(x)-1

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integral from 0 to x of tcos(t)

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

x sin (x) + cos (x) - 1

Solution steps

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

Apply Integration By Parts

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

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

\int sin (t) d t = - cos (t)

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

Simplify

= [t sin (t) + cos (t)]_{0}^{x}

Compute the boundaries:

x sin (x) + cos (x) - 1

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

\int_{0}^{1} x^{17} + 1 7^{x} d x

\int_{1}^{2} v d v

\int_{0}^{4} (x - 4) d x

\int_{0}^{5} (x^{2}) d x

\int_{0}^{1} 5^{5 x} d x

What is the integral from 0 to x of tcos(t) ?
The integral from 0 to x of tcos(t) is xsin(x)+cos(x)-1