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Help with this Integration please

  • Thread starter thaboy
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Summary:: I just need to know how we got from the 'beginning point' to the 'end point'/'answer'.

The left side is where we start and my professor did a bunch of calculations so fast that I wasn't able to understand how he got the result on the right side.

Could someone help me integrate this? I tried u-sub, trig, but had no luck so far. The x on the numerator is what's messing me up and I think the bounds too.

Please see attached image!
 

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PeroK
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Summary:: I just need to know how we got from the 'beginning point' to the 'end point'/'answer'.

The left side is where we start and my professor did a bunch of calculations so fast that I wasn't able to understand how he got the result on the right side.

Could someone help me integrate this? I tried u-sub, trig, but had no luck so far. The x on the numerator is what's messing me up and I think the bounds too.

Please see attached image!
What about ##y = x \sinh u##?
 
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What about ##y = x \sinh u##?
Perhaps, but my professor said it'll take no more than a simple Calc 1 integration and it's been a while since I've seen that kind of technique
 
PeroK
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Perhaps, but my professor said it'll take no more than a simple Calc 1 integration and it's been a while since I've seen that kind of technique
That substitution works and is a technique that every physics or maths student should be comfortable with.
 
SammyS
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Summary:: I just need to know how we got from the 'beginning point' to the 'end point'/'answer'.

The left side is where we start and my professor did a bunch of calculations so fast that I wasn't able to understand how he got the result on the right side.

Could someone help me integrate this? I tried u-sub, trig, but had no luck so far. The x on the numerator is what's messing me up and I think the bounds too.

Please see attached image!
From your image:
1575322737595.png


First of all: You do realize (I hope) that in this integral, y is the variable of integration, and x is to be treated as if it's a constant. Right?

So the indefinite integral associated with your expression is:

## \displaystyle \int \frac{a}{\left( a^2+u^2 \right) ^{3/2}} du ##

Use trig substitution or @PeroK's suggestion. No way do we know what your professor's method was.

For trig substitution: Set up a right triangle with acute angle ##\theta##. Seems obvious for the length of the hypotenuse to be ##\sqrt{a^2+u^2}## .
I chose ##u## as the length of the opposite leg, ##a## as length of adjacent leg.

I used ##\tan(\theta)## to get ##du##. The algebra can be worked out without too much difficulty. Using ##\cos(\theta)## instead, also works out with similar difficulty.

Notice that if you use ##\sin(\theta) = \dfrac{u}{\left( a^2+u^2 \right) ^{1/2}}## to get ##du##, you get a rather messy result for the derivative of the RHS. However, if you make the effort to simplify that messy result, you find that the derivative is nearly the same as the integrand, ##\dfrac{a}{\left( a^2+u^2 \right) ^{3/2}} ##. It's only different by a constant factor (multiple). In other words: this (the RHS, above) gives an anti-derivative without actually completing the trig substitution.
 

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