# Integral without using substitution method.

How to solve this integral without using substitution method? As I'm curious that is there another method to solve integral? I did integration by parts.

$$\int \sqrt{4-x^2} dx=x\sqrt{4-x^2}+\int \frac{x^2}{\sqrt{4-x^2}}dx$$ $$=x\sqrt{4-x^2}-\int \frac{4-x^2}{\sqrt{4-x^2}}dx+\int \frac{4}{\sqrt{4-x^2}}dx$$ $$=x\sqrt{4-x^2}-\int \sqrt{4-x^2}dx+4\sin ^{-1}(\frac{x}{2})$$ $$2\int\sqrt{4-x^2}dx=x\sqrt{4-x^2}+4\sin ^{-1}(\frac{x}{2})$$ $$\int \sqrt{4-x^2}dx=\frac{1}{2}x\sqrt{4-x^2}+2\sin ^{-1}(\frac{x}{2})$$

Is there any other method can solve integral other than substitution and this? I think Riemann Sum also can be used to solved. But people riemann sum is not considered a method of integration. I wonder why. Thanks a lot.

• By knowing what integrals of $\sqrt{a^2-x^2}$ are, allowed ;)? Mar 6 '15 at 15:00
• Please edit your question to include a description of what you've tried and where you're stuck. In this case, why are you disallowing substitutions? Mar 6 '15 at 15:00
• Why do you want to do this? If you had $\int_0^{2}\sqrt{4-x^2}\,\mathrm{d}x = \pi$, since it is part of a circle with radius $2$. Otherwise you need to do some work to figure out the integral, where substitution is the easiest. Mar 6 '15 at 15:00
• @N3buchadnezzar, actually there is a very nice geometric solution, fleshing out your observation about the definite integral, that avoids substitution. I would say more, but I think the OP needs to say more first. Mar 6 '15 at 15:22
• Yeah! I see it now as well using a drawing! clever However I think I will let my answer be until OP says more. Mar 6 '15 at 15:32

### Method 1: Integration by parts

$$\int \sqrt{a^2-x^2\,} \,\mathrm{d}x = \frac{x}{2}\sqrt{a^2-x^2\,} + \frac{a}{2} \int \frac{a}{\sqrt{a^2-x^2\,}\,}\mathrm{d}x \tag{1}$$

Pick $v' = x$ and $u = \sqrt{a^2-x^2}$ then solve with respect to $\int \sqrt{a^2-x^2\,} \,\mathrm{d}x$. Next part is to remember the derivative of the inverse sine function. $$\frac{\mathrm{d}}{\mathrm{d}x} \arcsin x = \frac{1}{\sqrt{1-x^2\,}\,} \tag{2}$$

Hence by using the chain rule, you can show that

$$\frac{\mathrm{d}}{\mathrm{d}x} \arcsin \left( \frac{x}{a} \right) = \frac{a}{\sqrt{a^2-x^2\,}\,} \tag{3}$$

Integrating $(3)$ w.r.t $x$ and then inserting the result into $(1)$ completes the calculations. I will leave it to you to fill in the details. Just ask if any part were particularly confusing.

### Method 2: Geometric considerations

here is a proof without words. I was able to discover this through the help of Barry Cipra in the comments above.

$$\int_0^x \sqrt{a^2-t^2\,} \,\mathrm{d}t = \color{blue}{\frac{x}{2}\sqrt{a^2-x^2\,}} + \color{green}{\frac{a^2}{2} \arcsin \left(\frac{x}{a}\right)}$$

The total area of a circle is $\pi a^2$. The area of the sector can be obtained by multiplying the circle's area by the ratio of the angle and $2 \pi$ (because the area of the sector is proportional to the angle, and $2 \pi$ is the angle for the whole circle, in radians): $$A = \pi a^2 \cdot \frac{\theta}{2 \pi} = \frac{a^2}{2} \theta$$ The rest follows since $\sin \theta = x/a$, where $a$ of course is the hypotenuse since it is the radius of the circle $x^2 + y^2 = a^2$. The area of the triangle should be straigt forward to figure out.

• what happens if $x < 0?$
– abel
Mar 8 '15 at 14:33
• you would then mirror the figure, and the calculations would be identical =) Mar 8 '15 at 14:35
• wont you get negative areas?
– abel
Mar 8 '15 at 14:38
• You are still above the x-axis. You would get $\int_0^{-x} \sqrt{a^2-t^2\,} \,\mathrm{d}t = \color{blue}{\frac{-x}{2}\sqrt{a^2-(-x)^2\,}} + \color{green}{\frac{a^2}{2} \arcsin \left(\frac{-x}{a}\right)}$ Because $x$ is now negative.. Is this suprising to you? Mar 8 '15 at 14:45
• if $x < 0,$ then you will need $\pi + \sin^{-1}(x)$ for the angle of the sector. it includes the quarter of a circle in addition to the triangle. you will have to argue much more carefully. but you can avoid all this if you keep $x \ge 0$ and deal with $x < 0$ by some sort of symmetry.
– abel
Mar 8 '15 at 14:50