What is the integral from-pi to 0 of sin(t)cos(nt) ?

The integral from-pi to 0 of sin(t)cos(nt) is -((-1)^n+1)/((n+1)(-n+1))

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integral from-pi to 0 of sin(t)cos(nt)

\int_{- π}^{0} sin (t) cos (n t) d t

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

Solution steps

\int_{- π}^{0} sin (t) cos (n t) d t

Use the following identity:

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

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

Take the constant out:

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

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

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_{- π}^{0} sin (t + n t) d t + \int_{- π}^{0} sin (t - n t) d t)

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

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

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

Simplify

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

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

\int_{- 1}^{1} (e^{x} - e^{2 x - 1}) d x

\int_{0}^{1} 2 t^{4} e^{t^{5}} d t

\int_{1}^{3} (- x^{2} + 4) d x

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

\int_{1}^{6} x^{3} d x

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