Which is the easiest way to evaluate ∫0π2(tan⁡x+cot⁡x) ?

Question

Durst37 · Accepted Answer

∫0π2tan⁡xdx+cot⁡xdx  =∫0π2sin⁡x+cos⁡xsin⁡xcos⁡xdx=∫0π2sin⁡x+cos⁡x2sin⁡xcos⁡x2dx=2  ∫0π2sin⁡x+cos⁡x1−(sin⁡x−cos⁡x)2dx  =2∫0π2sin⁡x+cos⁡x1−(sin⁡x−cos⁡x)2dx  Let t=sin⁡x−cos⁡x, dx=dtsin⁡x+cos⁡x  x→π2⇒t=(sin⁡x−cos⁡x)→1  x→0⇒t=(sin⁡x−cos⁡x)→−1  2∫−1111−t2dt=2[sin−1t]−11=2[π2−(−π2)]  =2π

Mollie Nash · Accepted Answer

Let u=tan⁡(x). Then u2=tan⁡(x) and 2udu=(1+tan2(x))dx. Thus  ∫0π2tan⁡(x)dx=∫0∞2u21+u4du  Since 1+u4=(1+2u+u2)(1−2u+u2), partial fraction decomposition applies:  2u21+u4=12(uu2−2u+1−uu2+2u+1)  Hence  ∫2u21+u4du=122log⁡(u2−2u+1u2+2u+1+tan−1(2+1)−tan−1(1−2u)2  Applying the fundamental theorem of calculus:  ∫0π2tan⁡(x)dx=π2

star233 · Accepted Answer

We will employing the substitution u=tan⁡x:u′=1+tan2⁡x2tan⁡xand2∫0π/2tan⁡xdx=4∫0∞u21+u4du=2∫−∞∞u21+u4duThe last integral has two poles (u1=eiπ/4, u2=ei3π/4) in the upper complex half-plane. The corresponding residue areResu=u1u21+u4=−u24 Resu=u1u21+u4=−u14Thus the value of the integral is2∫0π/2tan⁡xdx=−πi(u1+u2)=2π

Which is the easiest way to evaluate \int_0^{\pi/2}(\sqrt{\tam x}+\sqrt{\cot x})

Answered question

Answer & Explanation

New Questions in Calculus 1