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$$y = \int_\sqrt x^{\pi/4} \theta \tan\theta \, d\theta$$

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: https://www.google.ca/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwjC1fSiufTXAhUCU98KHRDnAWoQjRwIBw&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DIb568KmP5i0&psig=AOvVaw3FM9IGX9kUaHltIy3MqBDo&ust=1512617456071038

Michael Hardy
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Katelyn
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2 Answers2

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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.

  • I could do it if I was allowed to use the second part of the Fundamental Theorem of Calculus but I am only allowed to use the part of the Theorem that says that if a function f is continuous on [a,b] then g(x) is continuous on [a,b] and differentiable on (a,b) and g'(x)=f(x). – Katelyn Dec 06 '17 at 03:30
  • @Katelyn See my updated answer –  Dec 06 '17 at 03:41
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You have to use chain rule (of derivatives) combined with Fundamental Theorem of Calculus part 1. Put $u=\sqrt{x} $ and then we have $$\frac{d} {dx} \int_{\sqrt{x}} ^{\pi/4}\theta\tan\theta\, d\theta=\left(\frac {d} {du} \int_{u} ^{\pi/4}\theta \tan\theta\, d\theta\right)\frac{du} {dx} =(-u\tan u)\frac{1}{2\sqrt{x}}=-\frac{1}{2}\tan\sqrt{x} $$

Paramanand Singh
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