The value of the integral I = ∫(0 to π/4) cos(x) sin³(x) dx is?
Understand the Problem
The question is asking for the value of the integral, I = ∫(0 to π/4) cos(x) sin³(x) dx. This requires calculating the definite integral, which involves applying integration techniques.
Answer
The value of the integral is \( \frac{1}{16} \).
Answer for screen readers
The value of the integral ( I = \frac{1}{16} ).
Steps to Solve
- Substitution of Variables
We can simplify the integral by using the substitution $u = \sin(x)$. Therefore, $du = \cos(x) , dx$. The limits of integration change as follows:
- When $x = 0$, $u = \sin(0) = 0$.
- When $x = \frac{\pi}{4}$, $u = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.
So, the integral becomes:
$$ I = \int_{0}^{\frac{\sqrt{2}}{2}} u^3 , du $$
- Integrating the Function
Now, we can evaluate the integral:
$$ I = \int u^3 , du = \frac{u^4}{4} + C $$
- Apply the Limits of Integration
Next, we apply the limits of integration from $0$ to $\frac{\sqrt{2}}{2}$:
$$ I = \left[\frac{u^4}{4}\right]_{0}^{\frac{\sqrt{2}}{2}} $$
This results in:
$$ I = \frac{(\frac{\sqrt{2}}{2})^4}{4} - \frac{0^4}{4} $$
- Calculating the Final Expression
Now calculate:
$$ I = \frac{\left(\frac{2}{4}\right)^2}{4} $$
This simplifies to:
$$ I = \frac{\frac{1}{4}}{4} = \frac{1}{16} $$
The value of the integral ( I = \frac{1}{16} ).
More Information
This result shows that the definite integral of ( \cos(x) \sin^3(x) ) over the specified interval has a well-defined value. The technique of substitution is commonly used in integration to simplify complex integrals.
Tips
- Forgetting to Change Limits: Make sure to transform the limits of integration when substituting variables.
- Neglecting to Include the Constant of Integration: While calculating definite integrals, ensure you only apply the limits without a constant of integration.
- Misapplying Integration Formulas: Ensure that you are applying the correct power rule for integration.
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