While working on a subchapter in the chapter about integration named "The Fundamental theorem of Calculus", I am presented with the following task:

"Find the indicated derivative:

$$\frac{d}{d\theta}\int_{\sin \theta}^{\cos \theta} \frac{1}{1-x^2} dx$$"

I noticed that the task might get easier if I rewrote the expression as

$$\frac{d}{d\theta}(\int_{k}^{\cos \theta} \frac{1}{1-x^2} dx - \int_{k}^{\cos \theta} \frac{1}{1-x^2} dx)$$

I do not want to go through the trouble of actually integrating the expression, but rather use the fundamental theorem to draw some conclusions. I know that the integration will leave me with some expression including $\sin\theta$ and $\cos\theta$, and I was able to get this result down by intuition:

$$\frac{-\sin\theta}{1-\cos^2\theta} - \frac{\cos\theta}{1-\sin^2\theta}$$

Which, to my suprise, turned out to be correct. All I did was to add the derivative of the upper limit to the numerator of both expressions, while simply replacing $x$ in the denominator with the upper limit. I have some trouble seeing why this is correct, is there a solid, logical way to solve this without going through the trouble of doing the integration itself?

  • 1
    $\begingroup$ Chain rule...${}$ $\endgroup$ Oct 30, 2013 at 10:05
  • $\begingroup$ I do not see why you did not want to integrate and splitted the integral. The antiderivative is just (1/2) log[(1+x)/(1-x)]. Then, you can compute the values at the integration bounds and arrive to ArcTanh[Cos[t]] - ArcTanh[Sin[t]], which could still be simplified but not required since you need only the derivative with respect to t. Your result is correct and it can also be simplified to -(Csc[t]+Sec[t]). $\endgroup$ Oct 30, 2013 at 10:21

1 Answer 1


As @DavidMitra points out, this is the Chain rule : You set $$ F(y) = \int_k^y \frac{1}{1-x^2}dx $$ and you want to find $$ \frac{d}{d\theta}F(\cos(\theta)) $$ which, by the Chain rule and the Fundamental Theorem of Calculus, is $$ \frac{1}{1-\cos^2(\theta)}(-\sin(\theta)) $$ and the other integral is similar.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .