# Show that $\int_0^{\pi}\sqrt{9\sin^2t+16\cos^2t}\leq4\pi$

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.

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

Why not

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

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

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).

Doubly alternatively:

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