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For the following 3 questions, I know that the answer is false, but I'm not sure about how to provide a counter example. Can anyone please help me out?

Determine whether the following integrals are true or false. Provide a counter example if false and give an explanation if true:

a) If $x = 4\tan\theta$, then $\csc\theta = \frac{4}{x}$

b) The integral $\int_{1}^{2}\sqrt{x^2-1} dx$ does not have a real finite value

c) The integral $\int\frac{1}{x^2+4x+9} dx$ cannot be evaluated using trigonometric substitution

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  • $\begingroup$ A note on terminology: "true" and "false" are values given to statements. An integral can't be true or false any more than can a sum, because it doesn't assert. $\endgroup$
    – GFauxPas
    Mar 18, 2018 at 1:58

3 Answers 3

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a) This isn't really an integral here, but a (purported) identity. It claims that, given any $x$ and $\theta$ such that $x = 4\tan\theta$, then $\csc \theta = 4/x$ is also true. Make a counter-example by finding some specific value of $x$ and $\theta$ such that $x = 4\tan\theta$ is true, but $\csc \theta \neq 4/x$.

It shouldn't be too hard. Just choose a value of $\theta$, and let $x = 4 \tan \theta$. Most values of $\theta$ will produce a counter-example.

b) For this one, you can either compute the integral or just appeal to integration theorems. The fact that $\sqrt{x^2 - 1}$ is real-valued and continuous on $[1, 2]$ implies that it has some finite value. This is not strictly a counter-example though!

c) Provide a trigonometric substitution. A good start is to complete the square: $$\frac{1}{x^2 + 4x + 9} = \frac{1}{x^2 + 4x + 4 + 5} = \frac{1}{(x + 2)^2 + 5}.$$

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  • $\begingroup$ Alright, for b and c I can just provide a counter example by evaluating it? $\endgroup$
    – winonator
    Mar 18, 2018 at 1:35
  • $\begingroup$ Yes for b. For c, you need to evaluate it using a trig substitution. The question claims that no trig substitution will help. Prove it wrong by using one. $\endgroup$ Mar 18, 2018 at 1:38
  • $\begingroup$ Alright, I got it now, thanks. $\endgroup$
    – winonator
    Mar 18, 2018 at 1:39
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For b, notice that since $\sqrt{x^2-1}$ is monotonically increasing (from $1$ to $\infty$), and $\sqrt{2^2-1} < 2$, then $\int_1^2 \sqrt{x^2-1} dx < \int_1^2 2 dx$.

Therefore the area is less than $2$, and is definitely finite.

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a) If $ x = 4\tan\theta,$ then $\csc\theta = \frac 4x$ is not always true, because it implies $$ \tan\theta = \frac x4 = \sin\theta $$ which is not always true.

b) $$\int_{1}^{2}\sqrt{x^2-1} dx$$ is a real finite number because the integrand is continuous and the interval is bounded.

c) $$\int\frac{1}{x^2+4x+9} dx$$ can be evaluated using trigonometric substitution. complete the square and let $(x+2) = \sqrt 5 \tan (t)$

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