# An integration problem using trigonometric substitution

#### 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 $\sqrt{1+y^2}$.

• HappyS5

#### mech-eng

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.

#### HappyS5

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

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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.

• • HappyS5 and mech-eng

#### Orodruin

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I still recomment the hyperbolic substitution ......

• HappyS5

#### 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

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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.

#### 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. #### mech-eng

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

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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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