Question

In Exercises \(1-10,\) find the indefinite integral. $$\int 2 t \cos (3 t) d t$$

Short answer

The integral result is: \(\frac{2}{3} t \sin(3t) + \frac{2}{9}\cos(3t) + C\)

Step-by-step solution

Assign the functions

First identify the functions that are going to be \(u\) and \(dv\). A common way to choose \(u\) and \(dv\) for the purpose of integration by parts is to follow the mnemonic LIATE, which stands for Logarithm, Inverse trigonometric, Algebraic, Trigonometric and Exponential. The order represents the priority of selection as \(u\). Here, the algebraic function (which is \(t\)) is \(u\) and the remaining function (which is \(2 cos(3t)dt\)) would be \(dv\).\n Thus, \(u= t\) and \(dv= 2 \cos(3t)dt\).

Compute du and v

Next, we need to find the derivatives of \(u\) and integrals \(dv\). The derivative of \(t\) with respect to \(t\) gives \(du = dt\). The integral of \(2cos(3t)dt\) is \(\frac{2}{3}sin(3t)\) which will be our v. \n Thus, \(du= dt\) and \(v= \frac{2}{3}\sin(3t)\).

Apply the Integration by Parts formula

By the integration by parts formula, \(\int udv = uv - \int v du\), substitute our \(u\), \(v\), \(du\) and \(dv\) into the formula. This gives: \(t * \frac{2}{3}\sin(3t) - \int \frac{2}{3}\sin(3t) * dt\)

Compute the Remaining Integral

We now proceed to calculate the integral \(\int \frac{2}{3}\sin(3t) * dt = -\frac{2}{9}\cos (3t)\)

Combine and Simplify

Now combine everything giving the final solution of the integral: \(\frac{2}{3} t \sin(3t) + \frac{2}{9}\cos(3t) + C\), where \(C\) is the constant of integration.