Calculate the integral: integral from 0 to (pi/2) cos x

Calculate the integral: $\int_0^{\frac{\pi}{2}} \cos x \sin^2 x \, dx$.

We will use substitution to simplify this integral.

Step 1: Choose a substitution

Let $u = \sin x$. Then $du = \cos x \, dx$.

Step 2: Change the limits of integration

When $x = 0$: $u = \sin(0) = 0$. When $x = \frac{\pi}{2}$: $u = \sin\left(\frac{\pi}{2}\right) = 1$.

Step 3: Rewrite the integral

\[\int_0^{\frac{\pi}{2}} \cos x \sin^2 x \, dx = \int_0^1 u^2 \, du\]

Step 4: Integrate

\[\int_0^1 u^2 \, du = \left[\frac{u^3}{3}\right]_0^1\]

Step 5: Evaluate at the limits

\[= \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}\]

Verification using the antiderivative:

Since $u = \sin x$ gives us $F(x) = \frac{\sin^3 x}{3}$. Let's check: $F'(x) = \sin^2 x \cos x = \cos x \sin^2 x$. Applying the Newton-Leibniz formula:

\[\int_0^{\frac{\pi}{2}} \cos x \sin^2 x \, dx = \left[\frac{\sin^3 x}{3}\right]_0^{\frac{\pi}{2}} = \frac{1}{3} - 0 = \frac{1}{3}\]

This confirms our result!

Geometric interpretation:

The integral $\int_0^{\frac{\pi}{2}} \cos x \sin^2 x \, dx$ represents the area under the curve $y = \cos x \sin^2 x$ from $x = 0$ to $x = \frac{\pi}{2}$.

• At $x = 0$: $y = \cos(0) \sin^2(0) = 0$ (the curve starts at the origin)
• At $x = \frac{\pi}{2}$: $y = \cos\left(\frac{\pi}{2}\right) \sin^2\left(\frac{\pi}{2}\right) = 0$ (the curve ends at the $x$-axis)
• The function has a maximum somewhere in the interval. Taking the derivative: \[ \frac{d}{dx}[\cos x \sin^2 x] = -\sin^3 x + 2\sin x \cos^2 x = \sin x(2\cos^2 x - \sin^2 x) \] Setting this to zero (excluding $\sin x=0$): $2\cos^2 x = \sin^2 x = 1-\cos^2 x \Rightarrow 3\cos^2 x = 1 \Rightarrow \cos x = \frac{1}{\sqrt{3}}$. This occurs at $x = \arccos\left(\frac{1}{\sqrt{3}}\right) \approx 0.955$ radians (about $54.7°$).
• The function is positive throughout $(0, \frac{\pi}{2})$ since both $\cos x$ and $\sin^2 x$ are positive there.

General pattern:

This integral is a special case of the more general formula:

\[\int_0^{\frac{\pi}{2}} \cos x \sin^n x \, dx = \frac{1}{n+1}\]

For $n = 2$, we get $\frac{1}{3}$, which matches our result.

Comparison with related integrals:

Note the symmetry with the integral from Problem 32: $\int_0^{\pi/2} \sin x \cos^2 x \, dx = \frac{1}{3}$ (by symmetry or direct calculation) and $\int_0^{\pi/2} \cos x \sin^2 x \, dx = \frac{1}{3}$ (our current problem). This symmetry arises from the fact that sine and cosine are "swapped" over the interval $[0, \frac{\pi}{2}]$ under the transformation $x \to \frac{\pi}{2} - x$.

Final answer:

\[\boxed{\int_0^{\frac{\pi}{2}} \cos x \sin^2 x \, dx = \frac{1}{3}}\]