An integration problem using trigonometric substitution

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  • #26
Orodruin
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I am a bit confused. Why are people substituting ##3y = tan(\theta)##

I would substitute ##y=tan(\theta)##, and the deivative is ##\frac{dy}{dx} = tan^2(\theta) + 1 = sec^2(\theta)##. The final setup ## dy = sec^2(\theta) d \theta##.

##\int \sqrt(1 + y^2) dy = \int \sqrt(1 + tan^2(\theta)) \cdot sec^2(\theta) d\theta = \int sec^3(\theta) d\theta##

Now, I would use a reduction formula or integration by parts to solve the problem futher. The reduction formula can be derived by the integration by parts method.
The integrand is ##\sqrt{1+9y^2}##, not ##\sqrt{1+y^2}##.
 
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Now it seems that I can solve ##\sec^3\theta## hence I can solve the question. This is my result I have arrived at. Would you please check it?

##sec\theta=\frac {\sqrt {\frac13^2~+y^2}} {\frac13} ##

##\frac13\int sec^3\theta=\frac13 3\sqrt{{\frac13}^2+y^2}~~3y+\frac13 \frac12 ln |sec\theta+tan\theta|##

Substituting the values obtained and cancelling out,

##\frac13\int sec^3\theta=3\sqrt{\frac13^2+y^2}~~y+\frac16ln|\frac {\sqrt{\frac13^2+y^2}}{\frac13}|##

In the end, is this the solution?

P.S: When I asked this question, I thought this was a very straightforward and short question with a few easy steps.
 
  • #28
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The integrand is ##\sqrt{1+9y^2}##, not ##\sqrt{1+y^2}##.
Thanks.

In that case, I would solve it the following way:

##\int \sqrt{(1+(3y)^2)} dy \ and \ 3y = tan(\theta) \ and \ dy = \frac{sec^2(\theta) d\theta}{3}##

This gives ##\int \sqrt{(1 + tan^2(\theta))} \cdot \frac{sec^2(\theta)}{3} d\theta## or ## \frac {1}{3} \cdot \int sec^3(\theta) d\theta##

I would then solve the above by the reduction formula or integration by parts. As mentioned before, the reduction formula can be derived by integration by parts.
 
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  • #29
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##\frac13\int sec^3\theta=3\sqrt{\frac13^2+y^2}~~y+\frac16ln|\frac {\sqrt{\frac13^2+y^2}}{\frac13}|##

In the end, is this the solution?
Differentiate it and see if you recover what you started with.
 
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  • #30
Orodruin
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I still recomment the hyperbolic substitution ......
 
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  • #31
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Now it seems that I can solve ##\sec^3\theta## hence I can solve the question. This is my result I have arrived at. Would you please check it?

##sec\theta=\frac {\sqrt {\frac13^2~+y^2}} {\frac13} ##

##\frac13\int sec^3\theta=\frac13 3\sqrt{{\frac13}^2+y^2}~~3y+\frac13 \frac12 ln |sec\theta+tan\theta|##

Substituting the values obtained and cancelling out,

##\frac13\int sec^3\theta=3\sqrt{\frac13^2+y^2}~~y+\frac16ln|\frac {\sqrt{\frac13^2+y^2}}{\frac13}|##

In the end, is this the solution?

P.S: When I asked this question, I thought this was a very straightforward and short question with a few easy steps.
Remember the integration constant C.

Here is what I get:


##\frac{1}{3} \int sec^3(\theta) d\theta = \frac{1}{6} (9y \cdot \sqrt{(\frac{1}{3})^2 + y^2}## ##+ \ln \left|3 \big(\sqrt{(\frac{1}{3})^2 + y^2} + y) \right|)##+ C
 
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  • #32
Ray Vickson
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Now it seems that I can solve ##\sec^3\theta## hence I can solve the question. This is my result I have arrived at. Would you please check it?

##sec\theta=\frac {\sqrt {\frac13^2~+y^2}} {\frac13} ##

##\frac13\int sec^3\theta=\frac13 3\sqrt{{\frac13}^2+y^2}~~3y+\frac13 \frac12 ln |sec\theta+tan\theta|##

Substituting the values obtained and cancelling out,

##\frac13\int sec^3\theta=3\sqrt{\frac13^2+y^2}~~y+\frac16ln|\frac {\sqrt{\frac13^2+y^2}}{\frac13}|##

In the end, is this the solution?

P.S: When I asked this question, I thought this was a very straightforward and short question with a few easy steps.
You can check it for yourself: just differentiate the "answer" and see if it delivers the original integrand. This is something that you should get in the habit of doing, every time.
 
  • #33
ehild
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The original problem
##\int\sqrt{1+9y^2}dy##
can be solved without any substitution , just integrating by parts:
##\int{\sqrt{1+9y^2}(1)dy}=\sqrt{1+9y^2}y-\int{\frac {9y^2}{\sqrt{1+9y^2}}dy}##
Add and subtract 1 to the numerator of the fraction in the integral, it decomposes into two terms, one is the original integral, the other integral is a basic one. :smile:
 
  • #34
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The original problem
##\int\sqrt{1+9y^2}dy##
can be solved without any substitution , just integrating by parts:
##\int{\sqrt{1+9y^2}(1)dy}=\sqrt{1+9y^2}y-\int{\frac {9y^2}{\sqrt{1+9y^2}}dy}##
Add and subtract 1 to the numerator of the fraction in the integral, it decomposes into two terms, one is the original integral, the other integral is a basic one. :smile:
I don't understand what you do here mostly because there is no fraction in the original problem. It is just a square root of square of an expression of y.
 
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  • #35
ehild
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I don't understand what you do here mostly because there is no fraction in the original problem. It is just a square of an expression of y.
It is the integral in the second line.
 

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