What is the integral from 0 to pi of |sin(x)|cos(nx) ?

The integral from 0 to pi of |sin(x)|cos(nx) is -((-1)^{-n}+1)/((n+1)(n-1))

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integral from 0 to pi of |sin(x)|cos(nx)

\int_{0}^{π} ∣ sin (x) ∣ cos (n x) d x

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

Solution steps

\int_{0}^{π} ∣ sin (x) ∣ cos (n x) d x

Eliminate Absolutes

= \int_{0}^{π} sin (x) cos (n x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

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

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{π} sin (x + n x) + sin (x - n x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{π} sin (x + n x) d x + \int_{0}^{π} sin (x - n x) d x)

\int_{0}^{π} sin (x + n x) d x = \frac{( - 1 ) ^{n} + 1}{1 + n}

\int_{0}^{π} sin (x - n x) d x = \frac{( - 1 ) ^{- n} + 1}{1 - n}

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

Simplify

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

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

\int_{0}^{5} \frac{e ^{x}}{10 + e ^{x}} d x

\int_{0}^{2} k x^{3} d x

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

\int_{0}^{π} (sin (x))^{5} d x

\int_{0}^{1} e^{- s t} (- t) d t

What is the integral from 0 to pi of |sin(x)|cos(nx) ?
The integral from 0 to pi of |sin(x)|cos(nx) is -((-1)^{-n}+1)/((n+1)(n-1))