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I've been told that this integral admits a closed form $$ \int_0^{\Large\pi/3}\cosh^2\left(x/\sqrt{2}\right)\tan^3x \:dx$$ But an integration by parts with $u'(x)=\cosh^2\left(x/\sqrt{2}\right)$ and $v(x)=\tan^3x$ produces the factor $\tan^2x\sec^2x$ in my new integral... The other integration by parts doesn't seem that useful... Thanks for your help.

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Hint. Let's linearize $$ \cosh^2\left(x/\sqrt{2}\right)=\frac12\left( 1+\cosh \sqrt{2}x\right)$$ We have $$ \int_0^{\Large\pi/3}\cosh^2\left(x/\sqrt{2}\right)\tan^3x \:dx=\frac12\int_0^{\Large\pi/3}\tan^3x \:dx+\frac12\int_0^{\Large\pi/3}\cosh (\sqrt{2}x)\tan^3x \:dx$$ The first integral on the r.h.s. is easy, the second one reduces to evaluating $$ \int_0^{\Large\pi/3}e^{\sqrt{2}x}\tan^3(x) \:dx$$ and $$ \int_0^{\Large\pi/3}e^{-\sqrt{2}x}\tan^3(x) \:dx.$$ Integrating by parts twice, with $\displaystyle u'(x)=\tan^3(x)$ and $\displaystyle v(x)=e^{\sqrt{2}x}$ gives $$ \int e^{\sqrt{2}x}\tan^3(x) \:dx=\frac12 e^{\sqrt{2}x}(\sec^2 (x)-\sqrt{2}\tan(x))$$ implying $$ \int e^{-\sqrt{2}x}\tan^3(x) \:dx=\frac12 e^{-\sqrt{2}x}(\sec^2 (x)+\sqrt{2}\tan(x))$$ Then a closed form is deduced for your integral. Hoping you can take it from here.

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