How should i evaluate the following integral?
View attachment 11396
My teacher told me to use x = sec(t)/6 as an inverse trig substitution, but i end up with (secx)^3/216(tanx) and do not know how to proceed
Any hints?
I would advise you stay with MarkFL's posts
However, consider another way such as this:
Rewrite the integrand as:
\(\displaystyle x^2(36x^2 - 1)^{-1/2}(x \ dx) \)
Let \(\displaystyle \ u = 36x^2 - 1 \)
\(\displaystyle du = 72x \ dx \ \implies\)
\(\displaystyle x \ dx = \dfrac{1}{72}du\)
\(\displaystyle u = 36x^2 - 1 \ \implies \)
\(\displaystyle u + 1 = 36x^2 \ \implies \)
\(\displaystyle \dfrac{u + 1}{36} = x^2 \)
With \(\displaystyle \ u = 36x^2 - 1, \ \ \) when \(\displaystyle \ x = \dfrac{\sqrt{2}}{6}, \ \ u = 1\).
With \(\displaystyle \ u = 36x^2 - 1, \ \ \) when \(\displaystyle \ x = \dfrac{1}{6}, \ \ u = 0 \).
The integrand is \(\displaystyle \ \ \bigg(\dfrac{u + 1}{36}\bigg)\bigg(u^{-1/2}\bigg)\bigg(\dfrac{1}{72}du\bigg)\).
The integral is
\(\displaystyle \dfrac{1}{36(72)}\displaystyle\int_0^1(u + 1)u^{-1/2}du \ = \)
\(\displaystyle \dfrac{1}{2592}\displaystyle\int_0^1(u^{1/2} + u^{-1/2})du \)
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