What is the integral of t*cos(n*t*pi/2) ?

The integral of t*cos(n*t*pi/2) is 4/(pi^2n^2)(pi/2 ntsin(pi/2 nt)+cos(pi/2 nt))+C

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integral of tcos(nt*pi/2)

\int t \cdot cos (n \cdot t \cdot \frac{π}{2}) d t

\frac{4}{π ^{2} n ^{2}} (\frac{π}{2} n t sin (\frac{π}{2} n t) + cos (\frac{π}{2} n t)) + C

Solution steps

\int t cos (n t \frac{π}{2}) d t

\frac{π}{2} = π \frac{1}{2}

= \int t cos (n t π \frac{1}{2}) d t

Apply u-substitution

: \int \frac{4 u cos ( u )}{π ^{2} n ^{2}} d u

= \int \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 u cos (u) d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

= \frac{4}{π ^{2} n ^{2}} (u sin (u) - \int sin (u) d u)

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

= \frac{4}{π ^{2} n ^{2}} (u sin (u) - (- cos (u)))

Substitute back

u = n t π \frac{1}{2}

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

Simplify

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

= \frac{4}{π ^{2} n ^{2}} (\frac{π}{2} n t sin (\frac{π}{2} n t) + cos (\frac{π}{2} n t))

Add a constant to the solution

= \frac{4}{π ^{2} n ^{2}} (\frac{π}{2} n t sin (\frac{π}{2} n t) + cos (\frac{π}{2} n t)) + C

\frac{d}{d t} (v (t) e^{0.1 t})

\int x^{8} d x

\int_{0}^{1} x (x + 1) d x

\int \frac{e ^{0}}{( sin ( 3 x ) ) ^{2}} d x

n = 1 \sum \infty \frac{1}{2 ^{n} n}

What is the integral of t*cos(n*t*pi/2) ?
The integral of t*cos(n*t*pi/2) is 4/(pi^2n^2)(pi/2 ntsin(pi/2 nt)+cos(pi/2 nt))+C