The diagram shows part of the curve $y = 1 - \sin x$ passing through the point $P$. The curve touches the $x$-axis at the point $Q$. The tangent and normal to the curve at $P$ meet the $x$-axis at the points $R$ and $S$ respectively.

1. Set $y = 0$ since the curve touches the $x$-axis at $Q$.

$1 - \sin x = 0$

2. Solve the trigonometric equation for $x$.

$\sin x = 1$
$x = \frac{\pi}{2}$

1. Find the first and second derivatives of the curve

$y = 1 - \sin x$
$\frac{dy}{dx} = -\cos x$
$\frac{d^2y}{dx^2} = \sin x$

2. Locate the x-coordinate of P using the point of inflexion condition

At the point of inflexion $P$, $\dfrac{d^2y}{dx^2} = 0$:
$\sin x = 0$
For the relevant part of the curve (beyond $Q$ at $x = \tfrac{\pi}{2}$),
$x = \pi$

3. Find the y-coordinate of P

Substitute $x = \pi$ into $y = 1 - \sin x$:
$y = 1 - \sin \pi = 1 - 0 = 1$
Therefore $P = (\pi,\,1)$ (shown).

1. Find the derivative of the curve to determine the gradient function.

$y = 1 - \sin x$

$\frac{dy}{dx} = -\cos x$

2. Calculate the gradient of the tangent at point $P$.

At $P(\pi, 1)$, $x = \pi$.

Gradient of tangent, $m = -\cos(\pi) = -(-1) = 1$

3. Find the equation of the tangent at $P$.

Using $y - y_1 = m(x - x_1)$ with $P(\pi, 1)$ and $m = 1$:

$y - 1 = 1(x - \pi)$

$y = x - \pi + 1$

4. Find the $x$-coordinate of $R$ by setting $y = 0$.

Since $R$ lies on the $x$-axis, $y = 0$.

$0 = x - \pi + 1$

$x = \pi - 1$ (shown)

1. Find the gradient of the normal at $P$

From earlier parts, $P$ is $(\pi, 1)$ and $\frac{dy}{dx} = -\cos x$.
At $x = \pi$, the gradient of the tangent is $m_t = -\cos \pi = -(-1) = 1$.
The gradient of the normal is $m_n = -\frac{1}{m_t} = -1$.

2. Find the equation of the normal and its $x$-intercept

The equation of the normal at $P(\pi, 1)$ is:
$y - 1 = -1(x - \pi)$
$y = -x + \pi + 1$
At $S$, the normal meets the $x$-axis, so $y = 0$:
$0 = -x + \pi + 1$
$x = \pi + 1$ (shown)

1. Set up and integrate the curve from Q to P

Area under the curve from $Q\left(\tfrac{\pi}{2},0\right)$ to $P(\pi,1)$:
$\int_{\pi/2}^{\pi}(1-\sin x)\,dx = \big[\,x+\cos x\,\big]_{\pi/2}^{\pi}$

2. Evaluate the definite integral

$=\big(\pi+\cos\pi\big)-\left(\tfrac{\pi}{2}+\cos\tfrac{\pi}{2}\right)=(\pi-1)-\left(\tfrac{\pi}{2}+0\right)=\tfrac{\pi}{2}-1$

3. Find the area of the triangle under the normal

The normal meets the $x$-axis at $S(\pi+1,0)$. With $P(\pi,1)$ and foot $N(\pi,0)$:
$\text{base } NS=(\pi+1)-\pi=1,\qquad \text{height}=1$
$\text{Area of }\triangle PNS=\tfrac{1}{2}\times 1\times 1=\tfrac{1}{2}$

4. Combine the two areas

$\text{Shaded area}=\left(\tfrac{\pi}{2}-1\right)+\tfrac{1}{2}=\tfrac{\pi}{2}-\tfrac{1}{2}=\frac{\pi-1}{2}\ \text{units}^2$