3

I have this homework question in a first course of calculus.

enter image description here

The instruction ask to solve the integrals by substitute a=6 and b=6

My guess is to try to solve it by split the integral in 5 more simple ones.

enter image description here

I ask my teacher for some leads, he insist that there is no problem with the integral, also didn't tell me more. I need to know if it is possible to solve the integral without imaginary numbers and other issues that not concern to the first calculus semester? What do you think?

Haizum Skallah
  • 653
  • 3
    There is no need to do any integral. The $\sec^6(x)\tan^6(x)$ part of the integrand has two non-integrable singularities at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ in $[1,6]$ and nothing in the rest of the integrand that kills it. Furthermore, since the exponential of the $\int \sin^6(x)\cos^6(x)$ is positive and finite. If you insist to give a number to the integral, the number should be $\infty$. – achille hui Sep 13 '13 at 03:51
  • 1
    In B) the integrand is a complex-valued function. Taking $-1=i^2$ out the root, we obtain the real-valued integral wich can be calculated step by step with Maple 17 by $$with(Student[Calculus1]): IntTutor(12x/sqrt(x^2+6x-6), x = 1 .. 6);$$ See the output. – user64494 Sep 13 '13 at 04:15
  • @achillehui You are right! I didn't realize that it was a definite integral. – Matemáticos Chibchas Sep 13 '13 at 22:19
  • isn't there a missing $dx$? – zodiac Sep 15 '13 at 07:03
  • I think so, and also seems to me that there is a missing x in the cosine of the exponential integral (the second integral), Don't you think? – Haizum Skallah Sep 15 '13 at 14:23
  • @achillehui Put your comment as an answer, otherwise me or another user will undeservedly receive the bounty. – Matemáticos Chibchas Sep 16 '13 at 16:56

1 Answers1

4

Hint: If $u=\tan(x)$ then $du=\sec^2(x)\,dx$ and $\sec^{2m}(x)=(1+\tan^2(x))^m=(1+u^2)^m$.

Matemáticos Chibchas
  • 6,061