Question

i want to integrate this x*sqrt (a^2 - x^2)
let x= a sin z ; differentiating both side with respect to x , it becomes
1= a cos z * dz/dx
or dx = a cos z dz


if i differentiate both side with respect to z , it becomes

dx/dz =acosz
or dx = ac cos z dz

my question is which one is correct? differentiating with respect of x or with respect of z and why ?

Answer 1

Answer on Question #51672 – Math – Integral Calculus

I want to integrate this $\int x\sqrt{a^2 - x^2} dx$, let $x = a*\sin(z)$; differentiating both sides with respect to $x$, it becomes

$$1 = a^*\cos(z) \frac{dz}{dx} \quad \text{or} \quad dx = a^*\cos(z) \frac{dz}{dx}.$$

If I differentiate both sides with respect to $z$, it becomes

$$\frac{dx}{dz} = a^*\cos(z) \quad \text{or} \quad dx = a^*\cos(z) \frac{dz}{dzx}.$$

My question is which one is correct? Differentiating with respect to $x$ or with respect to $z$ and why?

Solution

Both versions are correct. Your purpose is to find $dx$ in form: $dx = f(z)dz$.

$$\frac{dx}{dz} = \frac{1}{\frac{dz}{dx}}.$$

So, it doesn't matter how to do differentiation: $dx/dz = f(z)$; $dz/dx = \frac{1}{f(z)}$.

From these two cases we obtain: $dx = f(z)dz$.

P.S.

Method 1

I show you an easier way:

$$\begin{aligned} \int x\sqrt{a^2 - x^2} dx &= |d(x^2)| = 2xdx| = \frac{1}{2} \int \sqrt{a^2 - x^2} d(x^2) = \frac{a}{2} \int \sqrt{1 - \frac{x^2}{a^2}} d(x^2) = \\ &- \frac{a^3}{2} \int \sqrt{1 - \frac{x^2}{a^2}} d\left(-\frac{x^2}{a^2}\right) = \\ &= - \frac{a^3}{2} \int \sqrt{1 - \frac{x^2}{a^2}} d\left(1 - \frac{x^2}{a^2}\right) = \left|1 - \frac{x^2}{a^2} = z\right| = - \frac{a^3}{2} \int z^{\frac{1}{2}} dz = - \frac{2a^3}{3 \cdot 2} z^{\frac{3}{2}} + C = - \frac{a^3}{3} \left(1 - \frac{x^2}{a^2}\right)^{\frac{3}{2}} + \\ &\quad C = - \frac{1}{3} (a^2 - x^2)^{\frac{3}{2}} + C, \end{aligned}$$

where $C$ is an arbitrary real constant.

Method 2

This method is a straightforward consequence of your question.

Assume that $\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2z} = \sqrt{a^2\cos^2z} = a|\cos z| = a \cdot \cos z$. Then

$$\int x \sqrt {a ^ {2} - x ^ {2}} d x = | x = a \cdot \sin z, d x = a \cdot \cos z d z | = \int a \cdot \sin z \cdot a ^ {2} \frac {\cos^ {2} z d z}{2} = - \frac {a ^ {3}}{2} \int \cos^ {2} z \cdot$$ $$(- \sin z d z) = - \frac {a ^ {3}}{2} \int \cos^ {2} z d (\cos z) = - \frac {a ^ {3}}{2} \frac {\cos^ {3} z}{3} + C = - \frac {a ^ {3}}{3} \left(1 - \frac {x ^ {2}}{a ^ {2}}\right) ^ {\frac {3}{2}} + C = - \frac {1}{3} (a ^ {2} - x ^ {2}) ^ {\frac {3}{2}} +$$ $$C,$$

where $C$ is an arbitrary real constant.

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