How do you integrate #g(x)=sinx/x^2# using the quotient rule?

Question

Ezekiel Alexander · Accepted Answer

See explanation.

From the topic you wrote in I assume you mean "how to differentiate the function?" The Quotient Rule says that if #f(x)# and #g(x)# are continuous functions, then to calculate the derivative #[(f(x))/g(x)]'# you can use the following formula: #[(f(x))/g(x)]'=(f'(x)g(x)-f(x)g'(x))/g^2(x)# For the given function the derivative is: #[sinx/x^2]'=((sinx)'*x^2-(x^2)'sinx)/x^4=(cosx*x^2-2xsinx)/x^4=(xcosx-x2sinx)/x^4=# #= (cosx-2sinx)/x^3#

Emily Ammerman · Answer

undefined You integrate $ g(x) = \frac{\sin(x)}{x^2} $ using integration by parts, not the quotient rule.

Scarlett Allison · Answer

undefined To integrate $ g(x) = \frac{\sin(x)}{x^2} $ using the quotient rule, we first rewrite the integral as:

$$ \int \frac{\sin(x)}{x^2} \, dx $$

Now, applying the quotient rule, we have:

$$ \int \frac{u}{v} \, dx = \int u \cdot \frac{1}{v} \, dx $$

Where $ u = \sin(x) $ and $ v = x^2 $.

Next, we need to find $ du $ and $ dv $:

$$ du = \frac{d(\sin(x))}{dx} = \cos(x) \, dx $$

$$ dv = \frac{d(x^2)}{dx} = 2x \, dx $$

Now, we apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substituting the values:

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

Simplify the integral:

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

Now, we have the integral of $ \frac{\cos(x)}{x} $, which does not have a simple closed-form antiderivative and is known as the sine integral function $ \text{Si}(x) $. So, we can rewrite it as:

$$ \int \frac{\cos(x)}{x} \, dx = \text{Si}(x) + C $$

Where $ C $ is the constant of integration.

Therefore, the final result of integrating $ g(x) = \frac{\sin(x)}{x^2} $ using the quotient rule is:

$$ \int \frac{\sin(x)}{x^2} \, dx = -\frac{\cos(x)}{x} + 2 \text{Si}(x) + C $$

Where $ C $ is the constant of integration.