$\begingroup$$$y = \int_\sqrt x^{\pi/4} \theta \tan\theta \, d\theta$$
I'm using the property of definite integrals that says $\int_b^a f(x) \, dx = -\int_a^b f(x) \, dx$
and I'm getting $y'= -\sqrt x \tan\sqrt x $ but the answer is $y'= -1/2 \tan\sqrt x $.
This is the Part 1 of the theorem:
$\endgroup$2 Answers
$\begingroup$You have to use chain rule (of derivatives) combined with Fundamental Theorem of Calculus part 1. Put $u=\sqrt{x} $ and then we have $$\frac{d} {dx} \int_{\sqrt{x}} ^{\pi/4}\theta\tan\theta\, d\theta=\left(\frac {d} {du} \int_{u} ^{\pi/4}\theta \tan\theta\, d\theta\right)\frac{du} {dx} =(-u\tan u)\frac{1}{2\sqrt{x}}=-\frac{1}{2}\tan\sqrt{x} $$
$\endgroup$$\begingroup$Change the variable $ \theta =\sqrt t$, $\frac {1}{2\sqrt t} dt = d\theta$ therefore $y(x) = \int_x^{(\frac {\pi}{4})^2}g(t)dt=\int_{(\frac {\pi}{4})^2}^x (-g(t))dt$. Should be easy from here.
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