ordinary differential equations exercise

Question

I'd like to resolve the ode $x'(t)=\cos(\ln(1+x(t)^2))$, given the inicial value $x(0)=1$, for t belonging to the interval [0,pi]. By separated variables method I get t equal to the primitive of the function $\sec(\ln(1+x^2))dx$, that I don't know how to calculate. Thank you very much in advanced!!

Answer 1

$x'(t)=\cos(\ln(1+x(t)^2))$

$\dfrac{dx}{dt}=\cos(\ln(1+x^2))$

$dt=\sec(\ln(1+x^2))~dx$

$\int_0^tdt=\int_1^x\sec(\ln(1+x^2))~dx$

$[t]_0^t=\int_1^x\sec(\ln(1+x^2))~dx$

$t=\int_1^x\sec(\ln(1+x^2))~dx$