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Sagot :
Answer:
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{4x-sin(4x)}{16} +c[/tex]
Step-by-step explanation:
Given
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx[/tex]
Required
Evaluate
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx[/tex]
Rewrite as:
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \int\limits {cos(x)\ sin(x)\ sin(2x)} \, dx[/tex]
In trigonometry:
[tex]sin(2x) = 2\ sin(x)\ cos(x)[/tex]
Divide both sides by 2
[tex]\frac{1}{2}sin(2x) = \frac{2\ sin(x)\ cos(x) }{2}[/tex]
[tex]\frac{1}{2}sin(2x) = sin(x)\ cos(x)[/tex]
[tex]\frac{1}{2}sin(2x) = cos(x)\ sin(x)[/tex]
Substitute [tex]\frac{1}{2}sin(2x)[/tex] for [tex]cos(x)\ sin(x)[/tex]
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \int\limits {\frac{1}{2}sin(2x)\ sin(2x)} \, dx[/tex]
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \int\limits {\frac{1}{2}sin^2(2x)} \, dx[/tex]
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{2}\int\limits {sin^2(2x)} \, dx[/tex]
Let [tex]u = 2x[/tex]
Differentiate:
[tex]du = 2 \ dx[/tex]
Make [tex]dx[/tex] the subject
[tex]dx = \frac{1}{2}du[/tex]
Substitute [tex]\frac{1}{2}du[/tex] for [tex]dx[/tex]
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{2}\int\limits {sin^2(2x)} \, \frac{1}{2}du[/tex]
Substitute 2x for u
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{2}\int\limits {sin^2(u)} \, \frac{1}{2}du[/tex]
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{2}*\frac{1}{2}\int\limits {sin^2(u)} \, du[/tex]
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}\int\limits {sin^2(u)} \, du[/tex]
At this point, we apply the reduction formula:
Which is:
[tex]\int\limits {sin^n(u)} \, du = \frac{n-1}{2}\int\limits sin^{n-2}(u)\ du\ - \frac{cos(u)sin^{n-1}(u)}{n}\du[/tex]
Let n = 2; So, we have:
[tex]\int\limits {sin^2(u)} \, du = \frac{2-1}{2}\int\limits sin^{2-2}(u)\ du\ - \frac{cos(u)sin^{2-1}(u)}{2}\du[/tex]
[tex]\int\limits {sin^2(u)} \, du = \frac{2-1}{2}\int\limits sin^{0}(u)\ du\ - \frac{cos(u)sin^{2-1}(u)}{2}\du[/tex]
[tex]\int\limits {sin^2(u)} \, du = \frac{1}{2}\int\limits sin^{0}(u)\ du\ - \frac{cos(u)sin^{2-1}(u)}{2}\du[/tex]
[tex]sin^0(u) = 1[/tex]
So, we have:
[tex]\int\limits {sin^2(u)} \, du = \frac{1}{2}\int\limits 1\ du\ - \frac{cos(u)sin^{2-1}(u)}{2}\du[/tex]
Integrate 1 with respect to u
[tex]\int\limits {sin^2(u)} \, du = \frac{1}{2}u - \frac{cos(u)sin^{2-1}(u)}{2}\du[/tex]
[tex]\int\limits {sin^2(u)} \, du = \frac{1}{2}u - \frac{cos(u)sin(u)}{2}\du[/tex]
Recall that:
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}\int\limits {sin^2(u)} \, du[/tex]
So, we have:
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}[ \frac{1}{2}u - \frac{cos(u)sin(u)}{2}\du][/tex]
Open bracket
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{8}u - \frac{cos(u)sin(u)}{8}[/tex]
Recall that: [tex]u = 2x[/tex] and [tex]du = 2 \ dx[/tex] [tex]dx = \frac{1}{2}du[/tex]
So, the expression becomes:
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{8}2x - \frac{cos(2x)sin(2x)}{8}[/tex]
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}x - \frac{cos(2x)sin(2x)}{8}[/tex]
Add constant c
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}x - \frac{cos(2x)sin(2x)}{8} +c[/tex]
----------------------------------------------------------------------------------------
In trigonometry:
[tex]sin(2\theta) = 2sin(\theta)cos(\theta)[/tex]
Divide both sides by 2
[tex]\frac{1}{2}sin(2\theta) = \frac{2sin(\theta)cos(\theta)}{2}[/tex]
[tex]\frac{1}{2}sin(2\theta) = sin(\theta)cos(\theta)[/tex]
Replace 2x with [tex]\theta[/tex]
[tex]\frac{1}{2}sin(2*2x) = sin(2x)cos(2x)[/tex]
[tex]\frac{1}{2}sin(4x) = sin(2x)cos(2x)[/tex]
----------------------------------------------------------------------------------------
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}x - \frac{cos(2x)sin(2x)}{8} +c[/tex] becomes
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}x - \frac{sin(4x)}{2*8} +c[/tex]
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}x - \frac{sin(4x)}{16} +c[/tex]
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{x}{4} - \frac{sin(4x)}{16} +c[/tex]
The solution can be further simplified as:
[tex]\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{4x-sin(4x)}{16} +c[/tex]
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