# Thread: Integration by inspection

1. ## Integration by inspection

The question:
Evaluate the following integral by inspection
$\displaystyle \int{\frac{sin(\sqrt{x})}{\sqrt{x}}}dx$

How would one solve this by inspection? The only thing I can think of is some sort of application of the product rule. But that doesn't sound like a process you'd do by inspection. :/

2. I'm not certain what "by inspection" means but I suspect it means there is an obvious substitution. You would be able to do it easily if it were only $\displaystyle \int sin(y)dy$, wouldn't you? Okay, make that happen!

3. I finally worked it out! Integration by substitution was further into the chapter. Thanks!

4. Originally Posted by Glitch
The question:
Evaluate the following integral by inspection
$\displaystyle \int{\frac{sin(\sqrt{x})}{\sqrt{x}}}dx$

How would one solve this by inspection? The only thing I can think of is some sort of application of the product rule. But that doesn't sound like a process you'd do by inspection. :/
No you dont use substitution.

You should recall that: $\displaystyle df(x)=f'(x)dx$ and if $\displaystyle f(x)=\sqrt{x}$ you actually have $\displaystyle d(\sqrt{x})=\frac{1}{2\sqrt{x}}dx$ which you can rewrite and use in that integral in the form of $\displaystyle \frac{dx}{\sqrt{x}}=2d(\sqrt{x})$.

$\displaystyle \int{\frac{sin(\sqrt{x})}{\sqrt{x}}}dx=\int\sin{(\ sqrt{x})}\frac{dx}{\sqrt{x}}=\int\sin (\sqrt{x})2d(\sqrt{x})=2\int\sin (\sqrt{x})d(\sqrt{x})=-2\cos(\sqrt{x})+C.$

You should always bear in mind that you could (and should be able to) use differential calculus just as you would use algebraic manipulation in order to simplify the integral.

5. Originally Posted by MathoMan
No you dont use substitution.

You should recall that: $\displaystyle df(x)=f'(x)dx$ and if $\displaystyle f(x)=\sqrt{x}$ you actually have $\displaystyle d(\sqrt{x})=\frac{1}{2\sqrt{x}}dx$ which you can rewrite and use in that integral in the form of $\displaystyle \frac{dx}{\sqrt{x}}=2d(\sqrt{x})$.

$\displaystyle \int{\frac{sin(\sqrt{x})}{\sqrt{x}}}dx=\int\sin{(\ sqrt{x})}\frac{dx}{\sqrt{x}}=\int\sin (\sqrt{x})2d(\sqrt{x})=2\int\sin (\sqrt{x})d(\sqrt{x})=-2\cos(\sqrt{x})+C.$

You should always bear in mind that you could (and should be able to) use differential calculus just as you would use algebraic manipulation in order to simplify the integral.
What you have just shown IS the method of substitution...

6. Originally Posted by Glitch
The question:
Evaluate the following integral by inspection
$\displaystyle \int{\frac{sin(\sqrt{x})}{\sqrt{x}}}dx$

How would one solve this by inspection? The only thing I can think of is some sort of application of the product rule. But that doesn't sound like a process you'd do by inspection. :/
My first thought upon seeing that integral was: chain rule. I know it may seem obvious, but it certainly helps to remind yourself that integration is simple the opposite of differentiation. I suppose what you mean by "integration by inspection" is to find the integral intuituively, and you would do so by asking yourself, "what can I differentiate to arrive at the integrand?" Using our knowledge of the chain rule, you would then notice that for $\displaystyle y=\sqrt{x}$, $\displaystyle \frac{dy}{dx}=\frac{1}{2\sqrt{x}}$.

7. Originally Posted by Prove It
What you have just shown IS the method of substitution...
I think not, but we can have different opinions, right! Let me clarify my stand.

What I demonstrated is a 'technique' sometimes called 'introducing terms under the differentiation sign' and substitution is a method of performing that 'technique' when terms are too complicated to 'introduce them under the differentiation sign' without abbreviating the notation by introducing a new variable. Only when you actually introduce a new variable (by denoting the substitution) you can speak of the method of substitution.

The biggest difference is when calculating definite integrals. When using the substitution method one has to ajust the limits of the integration according to the substitution used, while when performing the 'introducing terms under the differentiation sign' technique you are always in terms of the original variable (most often called x) and thus the limits of the integration remain the same.

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# inspection method in differentiation

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