How to GCF ( (Cosx)^4 - (Sinx)^4 )^1/2

Question

its actually a part of a question which is

finding indefinite integrals

pie/4

S  (1+(tanx)^2) / lsecxl ( (cosx)^4 - (sinx)^4)^1/2

-pie/4

Answer 1

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.