Question

Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) cos(x) sin(2x) sin(x) dx

Answer 1

Answer:

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{4x-sin(4x)}{16} +c$

Step-by-step explanation:

Given

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx$

Required

Evaluate

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx$

Rewrite as:

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \int\limits {cos(x)\ sin(x)\ sin(2x)} \, dx$

In trigonometry:

$sin(2x) = 2\ sin(x)\ cos(x)$

Divide both sides by 2

$\frac{1}{2}sin(2x) = \frac{2\ sin(x)\ cos(x) }{2}$

$\frac{1}{2}sin(2x) = sin(x)\ cos(x)$

$\frac{1}{2}sin(2x) = cos(x)\ sin(x)$

Substitute $\frac{1}{2}sin(2x)$ for $cos(x)\ sin(x)$

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \int\limits {\frac{1}{2}sin(2x)\ sin(2x)} \, dx$

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \int\limits {\frac{1}{2}sin^2(2x)} \, dx$

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{2}\int\limits {sin^2(2x)} \, dx$

Let $u = 2x$

Differentiate:

$du = 2 \ dx$

Make $dx$ the subject

$dx = \frac{1}{2}du$

Substitute $\frac{1}{2}du$ for $dx$

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{2}\int\limits {sin^2(2x)} \, \frac{1}{2}du$

Substitute 2x for u

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{2}\int\limits {sin^2(u)} \, \frac{1}{2}du$

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{2}*\frac{1}{2}\int\limits {sin^2(u)} \, du$

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}\int\limits {sin^2(u)} \, du$

At this point, we apply the reduction formula:

Which is:

$\int\limits {sin^n(u)} \, du = \frac{n-1}{2}\int\limits sin^{n-2}(u)\ du\ - \frac{cos(u)sin^{n-1}(u)}{n}\du$

Let n = 2; So, we have:

$\int\limits {sin^2(u)} \, du = \frac{2-1}{2}\int\limits sin^{2-2}(u)\ du\ - \frac{cos(u)sin^{2-1}(u)}{2}\du$

$\int\limits {sin^2(u)} \, du = \frac{2-1}{2}\int\limits sin^{0}(u)\ du\ - \frac{cos(u)sin^{2-1}(u)}{2}\du$

$\int\limits {sin^2(u)} \, du = \frac{1}{2}\int\limits sin^{0}(u)\ du\ - \frac{cos(u)sin^{2-1}(u)}{2}\du$

$sin^0(u) = 1$

So, we have:

$\int\limits {sin^2(u)} \, du = \frac{1}{2}\int\limits 1\ du\ - \frac{cos(u)sin^{2-1}(u)}{2}\du$

Integrate 1 with respect to u

$\int\limits {sin^2(u)} \, du = \frac{1}{2}u - \frac{cos(u)sin^{2-1}(u)}{2}\du$

$\int\limits {sin^2(u)} \, du = \frac{1}{2}u - \frac{cos(u)sin(u)}{2}\du$

Recall that:

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}\int\limits {sin^2(u)} \, du$

So, we have:

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}[ \frac{1}{2}u - \frac{cos(u)sin(u)}{2}\du]$

Open bracket

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{8}u - \frac{cos(u)sin(u)}{8}$

Recall that:  $u = 2x$ and  $du = 2 \ dx$      $dx = \frac{1}{2}du$

So, the expression becomes:

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{8}2x - \frac{cos(2x)sin(2x)}{8}$

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}x - \frac{cos(2x)sin(2x)}{8}$

Add constant c

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}x - \frac{cos(2x)sin(2x)}{8} +c$

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In trigonometry:

$sin(2\theta) = 2sin(\theta)cos(\theta)$

Divide both sides by 2

$\frac{1}{2}sin(2\theta) = \frac{2sin(\theta)cos(\theta)}{2}$

$\frac{1}{2}sin(2\theta) = sin(\theta)cos(\theta)$

Replace 2x with $\theta$

$\frac{1}{2}sin(2*2x) = sin(2x)cos(2x)$

$\frac{1}{2}sin(4x) = sin(2x)cos(2x)$

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$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}x - \frac{cos(2x)sin(2x)}{8} +c$ becomes

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}x - \frac{sin(4x)}{2*8} +c$

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{1}{4}x - \frac{sin(4x)}{16} +c$

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{x}{4} - \frac{sin(4x)}{16} +c$

The solution can be further simplified as:

$\int\limits {cos(x)\ sin(2x)\ sin(x)} \, dx = \frac{4x-sin(4x)}{16} +c$