integration of root a^2 +x^2, x^2 -a^2, a^2 -x^2

Integration of root a^2 +x^2, x^2 -a^2, a^2 -x^2 is given as

$ \int \sqrt {a^2 – x^2} dx = \frac {1}{2} x \sqrt {a^2 – x^2} + \frac {1}{2} a^2 \sin^{-1} \frac {x}{a} + C$

$ \int \sqrt {a^2 + x^2} dx = \frac {1}{2} x \sqrt {a^2 + x^2} + \frac {1}{2} a^2 ln |x +\sqrt {a^2 + x^2}| + C$

$ \int \sqrt {x^2 -a ^2} dx = \frac {1}{2} x \sqrt {x^2 – a^2} – \frac {1}{2} a^2 ln |x +\sqrt {x^2 – a^2}| + C$

Proof of Integration of root a^2 – x^2

Here, we can use the trigonometric substitution:
$ x = a\sin(\theta) $
$ dx =a\cos(\theta) \, d\theta $
Therefore
$ \int \sqrt{a^2 – a^2\sin^2(\theta)} a\cos(\theta) \, d\theta $
$ =a^2 \int \cos^2(\theta) \, d\theta $
This can be integrated using the half-angle formula:
$ \int \frac{1 + \cos(2\theta)}{2} \, d\theta $
$ = a^2 \frac{\theta}{2} + \frac{\sin(2\theta)}{4} + C $
Now, reverting the substitution:
$ \sin^{-1}(x/a) = \theta $
$ 2\sin(\theta)\cos(\theta) = \sqrt{1-x^2/a^2} $
Therefore
$ = a^2 \frac{\sin^{-1}(x/a)}{2} + \frac{x \sqrt{a^2 – x^2}}{2} + C $

Proof of Integration of root a^2 + x^2

Let $I= \int \sqrt {a^2 + x^2} dx $

To integrate $\sqrt{a^2 + x^2}$ using integration by parts, we need to identify parts of the function that can be set as (u) and (dv) in the integration by parts formula:

\[
\int u \, dv = uv – \int v \, du
\]

We’ll set:

• $u = \sqrt{a^2 + x^2}$
• (dv = dx)

$ I = x \sqrt {a^2 + x^2} – \int \frac {x^2}{\sqrt {a^2 + x^2} } \;dx $

$I= x \sqrt {a^2 + x^2} – \int \frac {a^2 + x^2 -a^2}{\sqrt {a^2 + x^2} } \;dx $

$I=x \sqrt {a^2 + x^2} – \int \sqrt {a^2 + x^2} dx + \int \frac {a^2}{\sqrt {a^2 + x^2} } \;dx$

$I=x \sqrt {a^2 + x^2} – I + \int \frac {a^2}{\sqrt {a^2 + x^2} } \;dx$

$2I=x \sqrt {a^2 + x^2} + a^2 \int \frac {1}{\sqrt {a^2 + x^2} } \;dx$

Now lets calculate $\int \frac {1}{\sqrt {a^2 + x^2} } \;dx$

$\int \frac {1}{\sqrt {a^2 + x^2}} dx = ln |x + \sqrt {a^2 + x^2}| + C$
Proof
Put $x =a tan \theta$ then $dx= a sec^2 \theta d\theta$
Therefore
$\int \frac {1}{\sqrt {a^2 + x^2}} dx$
$=\int sec \theta d\theta$
$=ln |sec \theta + tan \theta| + C$
Now
$tan \theta = \frac {x}{a}$
$sec \theta = \sqrt {1 + tan^2 \theta } = \sqrt {1 + \frac {x^2}{a^2}} $
Therefore
$=ln | \frac {x}{a} + \sqrt {1 + \frac {x^2}{a^2}}| + C_1 = ln |x + \sqrt {a^2 + x^2}| – ln |a| +C_1 = ln |x + \sqrt {a^2 + x^2}| + C$

Therefore substituting back

$2I= x \sqrt {a^2 + x^2} + a^2 ln |x + \sqrt {a^2 + x^2}| + C$

$I= \frac {1}{2} x \sqrt {a^2 + x^2} + \frac {1}{2} a^2 ln |x +\sqrt {a^2 + x^2}| + C$

Proof of Integration of root x^2 – a^2

Let $I= \int \sqrt {x^2 -a^2 } dx $

To integrate $\sqrt{x^2 – a^2}$ using integration by parts, we need to identify parts of the function that can be set as (u) and (dv) in the integration by parts formula:

\[
\int u \, dv = uv – \int v \, du
\]

We’ll set:

• $u = \sqrt{x^2 -a^2}$
• (dv = dx)

$ I = x \sqrt { x^2 -a^2} – \int \frac {x^2}{\sqrt {x^2 -a^2} } \;dx $

$I= x \sqrt { x^2 -a^2} – \int \frac {x^2 -a^2 +a^2}{\sqrt {x^2 a-a^2} } \;dx $

$I= x \sqrt { x^2 -a^2} – \int \sqrt {x^2 a-a^2} \;dx – \int \frac {a^2}{\sqrt { x^2 a-^2} } \;dx$

$2I= x \sqrt { x^2 -a^2} – \int \frac {a^2}{\sqrt { x^2 a-^2} } \;dx$

Now lets calculate $\int \frac {1}{\sqrt {x^2 -a^2} } \;dx$

$\int \frac {1}{\sqrt {x^2 – a^2}} dx = ln |x + \sqrt {x^2 – a^2}| + C$
Proof
Put $x =a sec \theta$ then $dx= a sec \theta tan \theta d\theta $
Therefore
$\int \frac {1}{\sqrt {x^2 -a^2}} dx $
$=\int sec \theta d\theta $
$=ln |sec \theta + tan \theta| + C $
Now
$sec \theta = \frac {x}{a} $
$tan \theta = \sqrt {sec^2 \theta -1 } = \sqrt {\frac {x^2}{a^2} -1} $
Therefore
$=ln | \frac {x}{a} + \sqrt {\frac {x^2}{a^2} -1}| + C_1 = ln |x + \sqrt {x^2 – a^2}| – ln |a| +C_1 = ln |x + \sqrt {x^2 -a^2}| + C$

Therefore substituting back

$2I= x \sqrt {x^2 -a^2} – a^2 ln |x + \sqrt {a^2 + x^2}| + C$

$I=\frac {1}{2} x \sqrt {x^2 – a^2} – \frac {1}{2} a^2 ln |x +\sqrt {x^2 – a^2}| + C$