Triggy Integral

Firstly notice that the integrand is of the form $\frac{g'(x)}{g(x)}\;\mathrm dx$ with $g(x)=\sin x + \cos x$ .

Therefore $\int \frac{g'(x)}{g(x)}\;\mathrm dx=\int\frac1g \;\mathrm dg=\log \lvert g(x) \rvert=\log \lvert \sin x + \cos x \rvert$

So we have $f(t)=\log \lvert \sin t + \cos t \rvert - \log \lvert \sin(-t) + \cos(-t) \rvert = \log \left\lvert \frac{\sin t + \cos t}{\cos t - \sin t} \right\rvert$

Then using some trigonometric identities the functionand of the logarithm turns into

$\frac{\sin t + \cos t}{\cos t - \sin t} =\frac{1+\tan t}{1- \tan t}$

Then notice that it is the logarithmic form of $\mathrm{artanh}$ which finally transforms the result into $f(t)=2\;\mathrm{artanh}(\tan t)$ Finally use a calculator and round to 3 s.f. to get $\fbox{1.23}$