What is the integral from 0 to pi/2 of sin(2nx) ?

The integral from 0 to pi/2 of sin(2nx) is (-(-1)^n+1)/(2n)

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integral from 0 to pi/2 of sin(2nx)

\int_{0}^{\frac{π}{2}} sin (2 n x) d x

\frac{- ( - 1 ) ^{n} + 1}{2 n}

Solution steps

\int_{0}^{\frac{π}{2}} sin (2 n x) d x

Apply Integration By Parts

: [sin (2 n x) x - \int cos (2 n x) \cdot 2 n x d x]_{0}^{\frac{π}{2}}

= [sin (2 n x) x - \int cos (2 n x) \cdot 2 n x d x]_{0}^{\frac{π}{2}}

\int cos (2 n x) \cdot 2 n x d x = x sin (2 n x) + \frac{1}{2 n} cos (2 n x)

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

Expand

sin (2 n x) x - (x sin (2 n x) + \frac{1}{2 n} cos (2 n x)) : - \frac{1}{2 n} cos (2 n x)

= [- \frac{1}{2 n} cos (2 n x)]_{0}^{\frac{π}{2}}

Compute the boundaries:

- (- 1)^{n} \frac{1}{2 n} + \frac{1}{2 n}

= - (- 1)^{n} \frac{1}{2 n} + \frac{1}{2 n}

Simplify

= \frac{- ( - 1 ) ^{n} + 1}{2 n}

\int_{1}^{2} x e^{3 x^{2}} d x

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\int_{\frac{π}{2}}^{π} sin (x) - x d x

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

What is the integral from 0 to pi/2 of sin(2nx) ?
The integral from 0 to pi/2 of sin(2nx) is (-(-1)^n+1)/(2n)