An integration problem using trigonometric substitution

[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

 

[Ray Vickson]

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]

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.

 

[ehild]

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.