Evaluate the definite integral J=-¼π∫¼π[(1+tan2(x))/|sec(x)|(cos4(x)-sin4(x))½]dx.
(cos4(x)-sin4(x))½=√(cos4(x)-sin4(x))=√[(cos2(x)-sin2(x))(cos2(x)+sin2(x))]=
√(cos2(x)-sin2(x))=√cos(2x).
1+tan2(x)=sec2(x), so (1+tan2(x))/|sec(x)|=sec2(x)/|sec(x)|=sec(x).
(1) J=-¼π∫¼πsec(x)√cos(2x)dx or (2) J=-¼π∫¼πsec(x)dx/√cos(2x), depending on interpretation of the question. J is the area between the curve f(x)=sec(x)√cos(2x) and the x-axis. For (1) f(π/4)=0=f(-π/4); for (2) f(±π/4)→∞. It seems likely that (1) is the required integral. Here, f(x) is symmetrical about the vertical axis so J=20∫¼πsec(x)√cos(2x)dx. The next problem is to evaluate the definite integral.
cos(2x)=1-2sin2(x); sec2(x)cos(2x)=sec2(x)-2tan2(x);
sec(x)√cos(2x)=√(sec2(x)-2tan2(x)); sec2(x)≡1+tan2(x),
sec(x)√cos(2x)=√(1-tan2(x)), J=20∫¼π√(1-tan2(x))dx.
J=1.3013 approx (see below).
Let u=tan(x), then du=sec2(x)dx=(1+u2)dx, dx=du/(1+u2).
When x=0, u=0; when x=π/4, u=1.
J=20∫1√(1-u2)du/(1+u2).
Note that because of the square root, u cannot exceed 1 or be less than -1, so let u=sinθ, du=cosθdθ.
When u=0, θ=0; when u=1, θ=π/2.
J=20∫½πcos2θdθ/(1+sin2θ)=20∫½πdθ/(sec2θ+tan2θ)=20∫½πdθ/(1+2tan2θ).
Let t=tanθ, then dt=sec2θdθ=(1+t2)dθ, dθ=dt/(1+t2). t=0 when θ=0; t→∞ when θ=π/2.
J=20∫∞dt/[(1+t2)(1+2t2)]=20∫∞(1/(½+t2)-1/(1+t2))dt.
Standard integral ∫dz/(a2+z2)=(1/a)tan-1(z/a); therefore, z=t and 1/a=√2 and 1 for the two integrals:
J=2[√2tan-1(t√2)-tan-1(t)]0∞=2(π√2/2-π/2)=π(√2-1)=1.30129 approx (confirmed by numerical analysis).
Note that, because this is a definite integral, there was no need to back substitute to reach a function of x. However, a more general solution (indefinite integral) can be written by performing back substitutions.