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Evaluate :

$\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$

My approach : I multiplied both sides by $\sqrt{a+x}$ and after simplification it comes down to :

$\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx + \int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} dx$, let's denote them by $I_1$ and $I_2$ respectively.

$I_1$ can be easily solved to $a \sin ^{-1} \frac{x}{a}$.

If I write $I_2$ as $ \frac{-1}{2}\int_{0}^{a} \frac{-2x}{\sqrt{a^{2}-x^{2}}} dx$, I get the solution as $\frac{a}{2}(\pi+1)$ but the answer given is $\frac{a}{2}(\pi+2)$. Where did I go wrong?

user91500
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Mojo Jojo
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5 Answers5

1

$$I_2=\frac{-1}{2} \int_0^a \frac{d(a^2-x^2)}{\sqrt{a^2-x^2}}$$

let $a^2-x^2=t$. If $x=0,$ then $t=a^2$ and if $x=a,$ then $t=0$.

Hence $$I_2=\frac{-1}{2} \int_{a^2}^0 \frac{dt}{\sqrt{t}}$$ $$\Rightarrow I_2= \int_{0}^{a^2} \frac{dt}{2\sqrt{t}}$$ $$\Rightarrow I_2=a-0=a$$

mea43
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0

Hint: Put $x=a \cos 2 \theta$.

0

Let $a^2-x^2=t$, then $-2xdx=dt$. Thus $$ \int_0^a \frac{x}{\sqrt{a^2-x^2}}dx=-\frac{1}{2}\int_{a^2}^0 \frac{1}{\sqrt{t}}dt =\left[\sqrt{t}\right]_0^{a^2}=a. $$

choco_addicted
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0

So for your solution:

$$I_2 = \int_{0}^{a} \frac{x}{\sqrt{a^2 - x^2}} dx$$

Let $u = a^2 - x^2, du = -2x dx, dx = -\frac{1}{2x} du$ then it is clear that

$$ I_2 = - \frac{1}{2} \int \frac{1}{\sqrt{u}} du = - \sqrt{u} $$

Now evaluating back at $x$, $$ = - \sqrt{a^2 - x^2} | _0^a = - \sqrt{0} + \sqrt{a^2} = a$$

I would recommend against, the expression

$$ - \frac{1}{2} \int_{0}^{a} \frac{-2x}{\sqrt{a^2 - x^2 }} dx $$

Since it doesn't really make it any easier to integrate.But if we insist, again making the substitution $u = a^2 - x^2, dx = -\frac{1}{2x} du$ we get:

$$ - \frac{1}{2} \int \frac{1}{\sqrt{u}} dx $$

Which is evaluated the same way.

Sidharth Ghoshal
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0

Assume $a\in\mathbb{R}^+$

$$\int_{0}^{a}\frac{\sqrt{a+x}}{\sqrt{a-x}}\space\text{d}x=$$

Substitute $u=\sqrt{a+x}$ and $\text{d}u=\frac{1}{2\sqrt{a+x}}\space\text{d}x$.

THis gives a new lower bound $u=\sqrt{a+0}=\sqrt{a}$ and upper bound $u=\sqrt{a+a}=\sqrt{2a}$:

$$\int_{\sqrt{a}}^{\sqrt{2a}}\frac{2u^2}{\sqrt{2a-u^2}}\space\text{d}u=2\int_{\sqrt{a}}^{\sqrt{2a}}\frac{u^2}{\sqrt{2a-u^2}}\space\text{d}u=$$

Substitute $u=\sqrt{2a}\sin(s)$ and $\text{d}u=\sqrt{2a}\cos(s)\space\text{d}s$.

Then $\sqrt{2a-u^2}=\sqrt{2a-2a\sin^2(s)}=\sqrt{2a}\cos(s)$ and $s=\arcsin\left(\frac{u}{\sqrt{2a}}\right)$.

THis gives a new lower bound $s=\arcsin\left(\frac{\sqrt{a}}{\sqrt{2a}}\right)=\frac{\pi}{4}$ and upper bound $s=\arcsin\left(\frac{\sqrt{2a}}{\sqrt{2a}}\right)=\frac{\pi}{2}$:

$$4a\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin^2(s)\space\text{d}s=4a\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left[\frac{1}{2}-\frac{\cos(2s)}{2}\right]\space\text{d}s=$$ $$4a\left[\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}1\space\text{d}s-\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos(2s)\space\text{d}s\right]=\frac{a(2+\pi)}{2}$$

Jan Eerland
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