What is the integral from 0 to 1 of tcos(n(pi/2)t) ?

The integral from 0 to 1 of tcos(n(pi/2)t) is (2(pinsin((pin)/2)+2cos((pin)/2)-2))/(pi^2n^2)

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integral from 0 to 1 of tcos(n(pi/2)t)

\int_{0}^{1} t cos (n (\frac{π}{2}) t) d t

\frac{2 ( πn sin ( \frac{πn}{2} ) + 2 cos ( \frac{πn}{2} ) - 2 )}{π ^{2} n ^{2}}

Solution steps

\int_{0}^{1} t cos (n \frac{π}{2} t) d t

\frac{π}{2} = π \frac{1}{2}

= \int_{0}^{1} t cos (nπ \frac{1}{2} t) d t

Apply u-substitution

: \int_{0}^{\frac{πn}{2}} \frac{4 u cos ( u )}{π ^{2} n ^{2}} d u

= \int_{0}^{\frac{πn}{2}} \frac{4 u cos ( u )}{π ^{2} n ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{4}{π ^{2} n ^{2}} \cdot \int_{0}^{\frac{πn}{2}} u cos (u) d u

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{0}^{\frac{πn}{2}}

= \frac{4}{π ^{2} n ^{2}} [u sin (u) - \int sin (u) d u]_{0}^{\frac{πn}{2}}

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

= \frac{4}{π ^{2} n ^{2}} [u sin (u) - (- cos (u))]_{0}^{\frac{πn}{2}}

Simplify

= \frac{4}{π ^{2} n ^{2}} [u sin (u) + cos (u)]_{0}^{\frac{πn}{2}}

Compute the boundaries:

\frac{πn}{2} sin (\frac{πn}{2}) + cos (\frac{πn}{2}) - 1

= \frac{4}{π ^{2} n ^{2}} (\frac{πn}{2} sin (\frac{πn}{2}) + cos (\frac{πn}{2}) - 1)

Simplify

= \frac{2 ( πn sin ( \frac{πn}{2} ) + 2 cos ( \frac{πn}{2} ) - 2 )}{π ^{2} n ^{2}}

x \to 3 lim (\frac{x + 3}{x - 3})

x \to 0 + lim (x^{18} ln (x))

x \to 3 - lim (\frac{1}{x - 3})

f (x) = \frac{5}{3 x}

\int (- 0.07 x) d x

What is the integral from 0 to 1 of tcos(n(pi/2)t) ?
The integral from 0 to 1 of tcos(n(pi/2)t) is (2(pinsin((pin)/2)+2cos((pin)/2)-2))/(pi^2n^2)