How do you use the comparison test to test for convergence if #1 / (xcosx) dx# from 0 to #pi/2#? Calculus Tests of Convergence / Divergence Direct Comparison Test for Convergence of an Infinite Series 1 Answer Jim H Apr 14, 2015 Compare to #1/x^2#. For #x# close to #0#, we have #x < cosx# so #1/cosx < 1/x# and #1/(xcosx) < 1/x^2# #int_0^(pi/2) 1/x^2 dx# converges, so does #int_0^(pi/2) 1/(xcosx) dx# Answer link Related questions What is the Direct Comparison Test for Convergence of an Infinite Series? How do you use the direct comparison test for infinite series? How do you use the direct comparison test for improper integrals? How do you use the direct Comparison test on the infinite series #sum_(n=1)^oo5/(2n^2+4n+3)# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^ooln(n)/n# ? How do you use the direct Comparison test on the infinite series #sum_(n=2)^oon^3/(n^4-1)# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^oo9^n/(3+10^n)# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^oo(1+sin(n))/(5^n)# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^ooarctan(n)/(n^1.2)# ? How do you use basic comparison test to determine whether the given series converges or diverges... See all questions in Direct Comparison Test for Convergence of an Infinite Series Impact of this question 5695 views around the world You can reuse this answer Creative Commons License