- Thread starter Moistwh
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\(\displaystyle I=\frac{1}{6^4}\int_0^{\frac{\pi}{4}} \left(\tan^2(t)+1\right)\sec^2(t)\,dt\)

Let:

\(\displaystyle u=\tan(t)\implies du=\sec^2(t)\,dt\)

And we have:

\(\displaystyle I=\frac{1}{6^4}\int_0^{1} u^2+1\,du=\frac{1}{972}\)

I would advise you stay with MarkFL's postsHow 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?

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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