Today’s quiz focused on integrating products of powers of trig functions. In these situations, we see powers of sine and cosine multiplied, or similarly secant and tangent. In our first problem, we have \(\int \cos^2(t)\sin^5(t)\,dt\).
Today’s first quiz problem started with an indefinite integral along with specific choices for \(u\) and \(dv\). With \(u = t\), then \(du\) is the derivative: \(du = 1\,dt\). With \(dv = \cos(5t)\,dt\), then \(v\) is antiderivative: \(v = \int \cos(5t)\,dt\).
Today’s first quiz problem is a definite integral. It is important to consider what our final result should look like — in this case, as a definite integral, our result should be a real number (representing the area bounded by \(f(x)\), the \(x\)-bounds, and the \(x\)-axis).
The WeBWorK Q&A site is a place to ask and answer questions about your homework problems. HINT: To ask a question, start by logging in to your WeBWorK section, then click “Ask for Help” after any problem.
Recent Comments