What is the integral of sin(t)cos(nt) ?

The integral of sin(t)cos(nt) is 1/2 (-1/(1+n)cos(t+nt)-1/(1-n)cos(t-nt))+C

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integral of sin(t)cos(nt)

\int sin (t) cos (n t) d t

\frac{1}{2} (- \frac{1}{1 + n} cos (t + n t) - \frac{1}{1 - n} cos (t - n t)) + C

Solution steps

\int sin (t) cos (n t) d t

Use the following identity:

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

= \int \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 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 sin (t + n t) d t + \int sin (t - n t) d t)

\int sin (t + n t) d t = - \frac{1}{1 + n} cos (t + n t)

\int sin (t - n t) d t = - \frac{1}{1 - n} cos (t - n t)

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

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{1 + n} cos (t + n t) - \frac{1}{1 - n} cos (t - n t)) + C

y^{^{''}} - 3 y^{^{'}} = e^{3 t}

k = 1 \sum \infty \frac{5 + ( - 2 ) ^{k}}{4 ^{k + 1}}

\int \frac{sin ^{7} ( 12 x )}{cos ^{4} ( 12 x )} d x

f (x) = ln (\frac{7 ( x + 3 )}{x})

\int x^{2} e^{- \frac{x}{3}} d x

What is the integral of sin(t)cos(nt) ?
The integral of sin(t)cos(nt) is 1/2 (-1/(1+n)cos(t+nt)-1/(1-n)cos(t-nt))+C