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tatyana61 [14]
3 years ago
7

Evaluate ∫ sin(x) cos(x) dx by four methods.

Mathematics
1 answer:
Sliva [168]3 years ago
3 0

(a) Since <em>u</em> = cos(<em>x</em>) gives d<em>u</em> = -sin(<em>x</em>) d<em>x</em>, we have

∫ sin(<em>x</em>) cos(<em>x</em>) d<em>x</em> = - ∫ (-sin(<em>x</em>)) cos(<em>x</em>) d<em>x</em>

= - ∫ cos(<em>x</em>) d(cos(<em>x</em>))

= - ∫ <em>u</em> d<em>u</em>

= - 1/2 <em>u</em>² + C

= -1/2 cos²(<em>x</em>) + C

(b) Now <em>u</em> = sin(<em>x</em>) gives d<em>u</em> = cos(x) d<em>x</em>, so

∫ sin(<em>x</em>) cos(<em>x</em>) d<em>x</em> = ∫ sin(<em>x</em>) d(sin(<em>x</em>))

= ∫ <em>u</em> d<em>u</em>

= 1/2 <em>u</em>² + C

= 1/2 sin²(<em>x</em>) + C

(c) Since sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>), we have

∫ sin(<em>x</em>) cos(<em>x</em>) d<em>x</em> = 1/2 ∫ sin(2<em>x</em>) d<em>x</em>

Substitute <em>u</em> = 2<em>x</em>, so that d<em>u</em> = 2 d<em>x</em>, and

∫ sin(<em>x</em>) cos(<em>x</em>) d<em>x</em> = 1/2 ∫ sin(2<em>x</em>) d<em>x</em>

= 1/4 ∫ 2 sin(2<em>x</em>) d<em>x</em>

= 1/4 ∫ sin(<em>u</em>) d<em>u</em>

= -1/4 cos(<em>u</em>) + C

= -1/4 cos(2<em>x</em>) + C

(d) Integrate by parts, setting

<em>u</em> = sin(<em>x</em>)          ==>  d<em>u</em> = cos(<em>x</em>) d<em>x</em>

d<em>v</em> = cos(<em>x</em>) d<em>x</em>  ==>  <em>v</em> = sin(<em>x</em>)

Then

∫ sin(<em>x</em>) cos(<em>x</em>) d<em>x</em> = sin²(<em>x</em>) - ∫ sin(<em>x</em>) cos(<em>x</em>) d<em>x</em>

2 ∫ sin(<em>x</em>) cos(<em>x</em>) d<em>x</em> = sin²(<em>x</em>) + C

∫ sin(<em>x</em>) cos(<em>x</em>) d<em>x</em> = 1/2 sin²(<em>x</em>) + C

The solutions in (b) and (d) are identical, but all 4 are equivalent, and this follows from the identities,

sin²(<em>x</em>) + cos²(<em>x</em>) = 1

cos(2<em>x</em>) = cos²(<em>x</em>) - sin²(<em>x</em>)

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To solve this, we need to find the distance from the origin to the y coordinate value, x coordinate value, then use the Pythagorean Theorem.

The origin of a graph (center) has the coordinates (0,0), so this will be our other coordinates.

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Next, let's find the y coordinate distance change. We move from the y coordinate 0, to the y coordinate 4, <u>so we move 4 y units.</u>

Now that we have these two leg lengths, let's imagine this as a right triangle. Moving from the origin, we draw a line from (0,0), to (-8,0). Then, we draw a line from (-8,0) to (-8,4). Now, draw a direct line from (0,0), to (-8,4). We have the length of the first(8) and second(4) lines, and we need to find the third line length to find our answer. To do this, we use the Pythagorean Theorem, which states that a²+b²=c². This says that, in a right triangle, the square of the two shorter lengths equals the square of the longest length. The longest length is what we are solving for.

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