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

Calculate the integral: $\int_0^{\frac{\pi}{2}} \sin x(1+\cos^2 x) \, dx$.

We will use substitution to simplify this integral.

Step 1: Expand the integrand

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

Step 2: Evaluate the first integral

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

Step 3: Evaluate the second integral using substitution

For $\int_0^{\frac{\pi}{2}} \sin x \cos^2 x \, dx$, let $u = \cos x$, so $du = -\sin x \, dx$, i.e. $\sin x \, dx = -du$.

Step 4: Change the limits of integration

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

Step 5: Rewrite the integral

\[\int_0^{\frac{\pi}{2}} \sin x \cos^2 x \, dx = \int_1^0 u^2 \cdot (-du) = \int_0^1 u^2 \, du\]

(reversing the limits removes the negative sign)

Step 6: Integrate

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

Step 7: Combine the results

\[\int_0^{\frac{\pi}{2}} \sin x(1+\cos^2 x) \, dx = 1 + \frac{1}{3} = \frac{4}{3}\]

Alternative approach - Direct substitution:

We could also use substitution on the entire integral from the start. Let $u=\cos x$, so $du=-\sin x\,dx$; when $x=0$, $u=1$; when $x=\frac\pi2$, $u=0$:

\[\int_0^{\frac{\pi}{2}} \sin x(1+\cos^2 x) \, dx = \int_1^0 (1+u^2)(-du) = \int_0^1 (1+u^2) \, du = \left[u + \frac{u^3}{3}\right]_0^1 = \frac{4}{3}\]

This confirms our result!

Geometric interpretation:

The integral $\int_0^{\frac{\pi}{2}} \sin x(1+\cos^2 x) \, dx$ represents the area under the curve $y = \sin x(1+\cos^2 x)$ from $x = 0$ to $x = \frac{\pi}{2}$.

• At $x = 0$: $y = \sin(0)(1+\cos^2(0)) = 0$
• At $x = \frac{\pi}{2}$: $y = \sin\left(\frac{\pi}{2}\right)\left(1+\cos^2\left(\frac{\pi}{2}\right)\right) = 1$
• At $x = \frac{\pi}{4}$: $y = \frac{\sqrt{2}}{2}\left(1+\frac{1}{2}\right) = \frac{3\sqrt{2}}{4}$
• The function is positive throughout $[0, \frac{\pi}{2}]$ and has a smooth curve

Verification using the antiderivative:

Since $u = \cos x$ gives us the antiderivative:

\[F(x) = -\cos x - \frac{\cos^3 x}{3}\]

Let's check: $F'(x) = \sin x + \sin x \cos^2 x = \sin x(1+\cos^2 x)$

\[\int_0^{\frac{\pi}{2}} \sin x(1+\cos^2 x) \, dx = \left[-\cos x - \frac{\cos^3 x}{3}\right]_0^{\frac{\pi}{2}} = 0 - \left(-1 - \frac{1}{3}\right) = \frac{4}{3}\]

Final answer:

\[\boxed{\int_0^{\frac{\pi}{2}} \sin x(1+\cos^2 x) \, dx = \frac{4}{3}}\]