- #26

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The integrand is ##\sqrt{1+9y^2}##, not ##\sqrt{1+y^2}##.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.