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How do you find the integral of #(sinx)(cosx)dx#?

Answer 1

Cameron Alexander

There are several ways to write the correct answer.

#intsinxcosxdx#

Solution 1 With #u = sinx# we get #sin^2x/2 +C#

Solution 2 With #u = cosx#, we get #-cos^2x/2 + c#

Solution 3 Noting that #sinxcosx = 1/2sin(2x)#, we rewrite:

#intsinxcosxdx = 1/2 int sin(2x) dx#

Now let #u = 2x# to get #1/4 cos(2x) + CC#

(I adore this problem and utilize it in all of my Calulus I lessons.)

Note 1: The fact that I represented the constants in each solution with a different notation is not by coincidence.

Note 2: The difference (literally -- that is, the subtraction) between the apparently different solutions are constants. For example: #sin^2x/2 # minus #-cos^2x/2# simplifies to:

#sin^2x/2 - (-cos^2x/2) = (sin^2x+cos^2)/2 = 1/2#

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Answer 2

Samantha Adkison

To find the integral of (sinx)(cosx)dx, you can use the substitution method. Let u = sin(x), then du = cos(x)dx. Rewrite the integral in terms of u: ∫ u du. Integrating u with respect to u gives u^2/2. Substitute back using u = sin(x), giving the final result of (sin(x))^2/2 + C, where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.