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Hello can you help me please with this problem about integrals, I don't know how to solve g(x), or, how can I find f '($π\over2$). Here is the problem Thanks in advance.

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Alexander
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2 Answers2

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Let $\;F,G\;$ be s.t. $\;F'=f\,,\,\,G'=g\;$ (primitive functions), then:

$$f(x)=F(g(x))-F(0)\;,\;\;g(x)=G(\cos x)-G(0)\implies$$

$$f'(x)=F'(g(x))\,g'(x)=f(g(x))\,g'(x)\;,\;\;g'(x)=G'(\cos x)\,(\cos x)'=g(\cos x)\,(-\sin x)$$

Now put all the above together, justify things and end the solution.

DonAntonio
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Note that $f(x)=F(g(x))$ where $$F(x)=\int_0^x \frac{1}{ \sqrt{1+t^3} }dt$$ and $g(x)=G(\cos x)$ where, again, $G$ is defined by $$G(x)=\int_0^x \big( 1+\sin (t^2) \big)dt$$ By the FTC we have that $$F'(x)=\frac{1}{ \sqrt{1+x^3} }$$ and that $G'(x)=1+\sin (x^2)$.

Now, by the chain rule, we have that $$g'(x)=G'(\cos x)\cos' x = -G'(\cos x)\sin x = -\big( 1+\sin (\cos^2 x) \big) \sin x$$ Therefore $$f'(\tfrac{\pi}{2})=F'(g(\tfrac{\pi}{2}))g'(\tfrac{\pi}{2})=\frac{1}{ \sqrt{1+(g(\tfrac{\pi}{2}))^3} } g'(\tfrac{\pi}{2})=\frac{1}{ \sqrt{1+0^3} } (-1)=-1$$

azif00
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  • Thank you friend, that's correct, -1 is the answer, but how do you know that g(π/2) is 0, solving without an calculator? – Alexander Aug 01 '19 at 22:18
  • $g(\tfrac{\pi}{2})=G(\cos \tfrac{\pi}{2})=G(0)$ and $G(0)$ is zero since $$\int_0^0 \big( 1+\sin (t^2) \big)dt=0$$ and in general $$\int_a^a f(x)dx=0$$ – azif00 Aug 01 '19 at 22:42
  • Now I understand. Thank you so much – Alexander Aug 01 '19 at 23:48