# An integration problem using trigonometric substitution

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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}##.

HappyS5
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.

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.

mech-eng
vela
Staff Emeritus
Homework Helper
##\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.

HappyS5 and mech-eng
Orodruin
Staff Emeritus
Homework Helper
Gold Member
I still recomment the hyperbolic substitution ......

Pi-is-3 and HappyS5
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

ehild and mech-eng
Ray Vickson
Homework Helper
Dearly Missed
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.

ehild
Homework Helper
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.

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.
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.

Last edited:
ehild
Homework Helper
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.