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I can't to solve it.

Since $\int_0^{\pi}\sqrt{9\sin^2t+16\cos^2t} dt=\int_0^{\pi}\sqrt{9+7\cos^2t}dt\leq\int_0^{\pi}[\sqrt{9}+\sqrt{7\cos^2t}]dt=3\pi+2\sqrt{7}$

But $4\pi\leq3\pi+2\sqrt{7}.$

Please help me to Show that $\int_0^{\pi}\sqrt{9\sin^2t+16\cos^2t}\leq4\pi$ without using the calculator.

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    $\begingroup$ $\cos^2 t \le 1$, then $\sqrt{9+7\cos ^2 t }\le 4$. $\endgroup$ Mar 8, 2020 at 18:26
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    $\begingroup$ Look like it is the polar coordinate of an ellipse, which has major axis =4 and minor 3.so area would be $$\pi ab$$. $\endgroup$
    – Jack Rod
    Mar 9, 2020 at 3:31
  • $\begingroup$ $$\int_0^{\pi}\sqrt{9\sin^2t+16\cos^2t}\, dt=\int_0^{\pi}\sqrt{16-7\sin^2t}\,dt\leq \int_0^{\pi}\sqrt{16}\,dt=4\pi$$ $\endgroup$
    – Axion004
    Mar 9, 2020 at 17:35

3 Answers 3

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Why not

$$\int_0^{\pi}\sqrt{9\sin^2t+16\cos^2t}\, dt=\int_0^{\pi}\sqrt{9+7\cos^2t}\,dt\leq \int_0^{\pi}\sqrt{9+7}\,dt=4\pi\ ?$$

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    $\begingroup$ +1 why be complicated when simple solutions exist. $\endgroup$ Mar 8, 2020 at 18:23
  • $\begingroup$ Ohhhh, I feel dumb. Thank you very much. $\endgroup$
    – Zera
    Mar 8, 2020 at 18:25
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    $\begingroup$ Can down voter explain us, why did you do it? +1 $\endgroup$ Mar 8, 2020 at 18:25
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    $\begingroup$ @MichaelRozenberg, you need to have a thick skin around here sometimes. I got used to the ocasional senseless downvote :) $\endgroup$
    – LHF
    Mar 8, 2020 at 18:29
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    $\begingroup$ It could easily have been a misclick - especially on mobile. I've done it more than a few times (and corrected it, of course). $\endgroup$ Mar 8, 2020 at 18:45
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Option:

$\sqrt{9\sin^2 t +16 \cos ^2 t } =$

$\sqrt{16 \sin^2 t +16 \cos^2 t -7 \sin^2 t} \le$

$\sqrt{16}=4$, since $\sin^2 t \ge 0.$

Note: $x=3\cos t$; $y=4 \sin t$ is an ellipse:

$x^2/3^2 +y^2/4^2=1$;

The integral is the arc length, $0\le t\le π$ (half the perimeter) of this ellipse.

Inequality: Half the perimeter of this ellipse is $\le 4π$, half the perimeter of the circle with radius $=4$ {lenght of the major axis of the ellipse).

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Doubly alternatively:

$$\sqrt{9\sin^2(t) + 16\cos^2(t)} \le \sqrt{16\sin^2(t)+16\cos^2(t)} = \sqrt{16} = 4.$$

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