Calculate the following antiderivative. Show your work. \int \sin(x) \cos(x) dx

Question:

Calculate the following antiderivative. Show your work. {eq}\int \sin(x) \cos(x) dx {/eq}

Indefinite Integral:

Assume that {eq}q {/eq} is a single-valued function of {eq}x {/eq}. If there exists a function {eq}H {/eq} is called an anti derivative of {eq}q {/eq} if we have {eq}\displaystyle q\left( x \right) = H'\left( x \right),\;\forall \;x {/eq} . The general form {eq}\displaystyle H(x)+c {/eq} (where {eq}c {/eq} is an integration constant) is an indefinite integral of {eq}\displaystyle q {/eq} and is denoted by {eq}\displaystyle \int q\left( x \right)dx = H\left( x \right) + C {/eq}.

The following rules are relevant to this problem:

1.{eq}{\int {\sin x} dx = - \cos x + c} {/eq}

2.{eq}{\sin 2x = 2\sin x\cos x} {/eq}

Answer and Explanation:

Given that: {eq}\displaystyle \int {\sin } (x)\cos (x)dx {/eq}

{eq}\displaystyle\ \eqalign{ & \int {\sin } (x)\cos (x)dx \cr & {\text{From identity multiplication;}} \cr & = \frac{1}{2}\int {2\sin } (x)\cos (x)dx \cr & {\text{From sine }}2x{\text{ form;}} \cr & = \frac{1}{2}\int {\sin 2x} dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sin 2x = 2\sin x\cos x} \right) \cr & = - \frac{1}{2}\left[ {\frac{{\cos 2x}}{2}} \right] + c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int {\sin x} dx = - \cos x + c} \right) \cr & = - \frac{1}{4}\cos 2x + c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {c{\text{ is integration constant}}{\text{.}}} \right) \cr & \cr & \int {\sin } (x)\cos (x)dx = - \frac{1}{4}\cos 2x + c \cr} {/eq}


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Indefinite Integral: Definition, Rules & Examples

from Calculus: Tutoring Solution

Chapter 7 / Lesson 14
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