The value of the integral I = ∫(0 to π/4) cos(x) sin³(x) dx is?

Steps to Solve

1. 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:

$$ I = \int_{0}^{\frac{\sqrt{2}}{2}} u^3 , du $$

2. Integrating the Function

Now, we can evaluate the integral:

$$ I = \int u^3 , du = \frac{u^4}{4} + C $$

3. 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:

$$ I = \frac{(\frac{\sqrt{2}}{2})^4}{4} - \frac{0^4}{4} $$

4. Calculating the Final Expression

Now calculate:

$$ I = \frac{\left(\frac{2}{4}\right)^2}{4} $$

This simplifies to:

$$ I = \frac{\frac{1}{4}}{4} = \frac{1}{16} $$