![]() The square root is a #1/2# power, which can be brought from the logarithm rule: log(A^B)=Blog(A)#. However, we can simplify it rather sneakily: To remedy this, add absolute value bars.) This is not the case in the original function, but the trigonometry introduced this issue. This almost could be a final answer, but there is one problem, which is that #sqrt(x^2-3)# in the answer restricts the domain, in that we see that the domain excludes #-sqrt3 < x < sqrt3#. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Thus: #" "csctheta=x/sqrt(x^2-3)" "#and #" "cottheta=sqrt3/sqrt(x^2-3)# By the Power Rule, the integral of x3 2 x 3 2 with respect to x x is 2 5x5 2 2 5 x 5 2. Note that #sectheta=x/sqrt3#, thus we have a right triangle where #x# is the hypotenuse, #sqrt3# is the side adjacent to #theta#, and the opposite side is #sqrt(x^2-3)#. #I=1/sqrt3int(secthetatanthetad theta)/tan^2theta=1/sqrt3int(secthetad theta)/tantheta=1/sqrt3int1/costheta(costheta/sintheta)d theta=1/sqrt3intcscthetad theta# #I=int(sqrt3secthetatanthetad theta)/(3sec^2theta-3)=sqrt3/3int(secthetatanthetad theta)/(sec^2theta-1)# Let #x=sqrt3sectheta#, implying that #dx=sqrt3secthetatanthetad theta#: ![]()
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