How can I find the area bounded by given curves?

Question

Find the area bounded by the curve and the axes whose parametric equations are $$ x=a\cos 2\theta, \quad y=b\sin 2\theta, $$ where $0\leqslant \theta\leqslant\pi$ and $0<a<b$, from $x=0$ to $x=a$.

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I've attempted it multiple times and I got $0.785ab$, whereas the given answer is $1.57ab$.

Answer 1

The standard parameterisation of ellipse of form $$\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$$

is $x = a \cos t$ and $ y = b \sin t$ for $0 \leq t \leq 2\pi$, the area of such a ellipse is $ab\pi$.

Let $ \theta = \dfrac{t}2$, where $ 0 \leq \theta \leq \pi$ and then $x = a \cos(2 \theta)$ and $ y = b\sin (2\theta)$.

We need to find the area from $x = 0 \left(\theta = \dfrac\pi2 \right)$ to $ x = a \left(\theta = 0 \right)$, which is a half of the ellipse. Hence the given area is $\dfrac\pi2ab$.

Answer 2

$$A=\int_0^\pi x(\theta)y'(\theta)d\theta$$ $$\frac{dy}{d\theta}=2b\cos(2\theta)$$ $$a\int_0^\pi \cos(2\theta)2b\cos(2\theta)d\theta$$ $$2ab\int_0^\pi \cos^2(2\theta)d\theta$$ Use $2\theta=u$ where $2d\theta=du$ $$ab\int \cos^2(u)du$$ $$\frac{ab}{2}\int \cos(2u)+1du$$ $$\frac{ab}{2}[ \frac{\sin(2u)}{2}+u]$$ $$\frac{ab}{2}[ \frac{\sin(4\theta)}{2}+2\theta]$$ From our initial integration limits apply $\theta$ on $\pi$ and $0$ $$\frac{ab}{2}[\frac{\sin(4\pi)}{2}+2\pi-\frac{\sin(0)}{2}-0]$$ Which leads to $ab\pi$ which is congruent to any circle or ellipse described by the parametric equations. I didnt understand the question but if you were to take the positive side of x or y you would get the response you needed