Question

find the antiderivative for each function when c equals 0. do as many as you can mentally. check your answers by differentiation.
a. -8 csc x cot x
b. - csc (6x) cot (6x)
c. -π csc (πx/4) cot (πx/4)
a. the antiderivative of -8 csc x cot x is 8 csc x.
b. the antiderivative of - csc (6x) cot (6x) is 1/6 csc (6x)
c. the antiderivative of -π csc (πx/4) cot (πx/4) is (type an exact answer, using π as needed.)

Explanation:

Step1: Recall the antiderivative formula

The antiderivative of \(a\csc(u)\cot(u)\) is \(-a\csc(u)\div u'\) (by the chain - rule for integration). For the function \(y = \csc(u)\), \(y'=-\csc(u)\cot(u)u'\).

Step2: Identify \(u\) and \(u'\) for the function \(y =-\pi\csc(\frac{\pi x}{4})\cot(\frac{\pi x}{4})\)

Let \(u=\frac{\pi x}{4}\), then \(u'=\frac{\pi}{4}\)

Step3: Apply the antiderivative formula

Using the formula \(\int a\csc(u)\cot(u)dx=-\frac{a}{u'}\csc(u)+C\) (with \(C = 0\) in our case). Here \(a =-\pi\) and \(u'=\frac{\pi}{4}\)

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Step4: Check by differentiation

Differentiate \(y = 4\csc(\frac{\pi x}{4})\) using the chain - rule. If \(y = 4\csc(u)\) with \(u=\frac{\pi x}{4}\), then \(y'=4(-\csc(u)\cot(u))\times\frac{\pi}{4}=-\pi\csc(\frac{\pi x}{4})\cot(\frac{\pi x}{4})\)

Answer:

\(4\csc(\frac{\pi x}{4})\)