Question

Admitere politehnica Timișoara ex:AM 216. sa se calculeze valoarea integralei​.
Integrala de la 0 la 1 din(tg^5 x+tg^7 x +tg^9 x + tg^11 x ) dx
Raspunsuri: a.-3/4 b.5/4 c.144/5 d.2/3 e.11/6 f.155/6

Answer 1

Salut,

$\displaystyle\int_{0}^{\frac{\pi}{3}}(tg^5x+tg^7x+tg^9x+tg^{11}x)dx=\int_{0}^\frac{\pi}{3}[tg^5x(1+tg^2x)+tg^9x(1+tg^2x)]dx=\\\\=\int_{0}^\frac{\pi}{3}(1+tg^2x)(tg^5x+tg^9x)dx=\int_{0}^\frac{\pi}{3}\left(1+\dfrac{sin^2x}{cos^2x}\right)(tg^5x+tg^9x)dx=\\\\=\int_{0}^\frac{\pi}{3}\dfrac{sin^2x+cos^2x}{cos^2x}\cdot(tg^5x+tg^9x)dx=\int_{0}^\frac{\pi}{3}\dfrac{1}{cos^2x}\cdot(tg^5x+tg^9x)dx=\\\\=\int_{0}^\frac{\pi}{3}(tgx)^{'}\cdot(tg^5x+tg^9x)dx=\int_{0}^\frac{\pi}{3}(tgx)^{'}\cdot tg^5xdx+\int_{0}^\frac{\pi}{3}(tgx)^{'}\cdot tg^9xdx=\\\\=\dfrac{tg^6x}6\Bigg{|}_0^{\frac{\pi}{3}}+\dfrac{tg^{10}x}{10}\Bigg{|}_0^{\frac{\pi}{3}}=\dfrac{1}6\left[tg^6\left(\frac{\pi}{3}\right)-tg^60\right]+\dfrac{1}{10}\left[tg^{10}\left(\frac{\pi}{3}\right)-tg^{10}0\right]=\\\\=\dfrac{1}6\left[(\sqrt3)^6-0^6\right]+\dfrac{1}{10}\left[(\sqrt3)^{10}-0^{10}\right]=\dfrac{3^3}6+\dfrac{3^5}{10}=\dfrac{3^2}2+\dfrac{3^5}{10}=\dfrac{144}{5}.$

Răspunsul corect este deci c.

Green eyes.

Answer 2

$I = \int_0^{\frac{\pi}{3}}(\tan^5x+\tan^7x+\tan^9x+\tan^{11}x)\,dx \\ \\ \tan x = t \Rightarrow \dfrac{1}{\cos^2 x} \, dx = dt\\ \\ I = \int_{0}^{\frac{\pi}{3}}(\tan^5x(1+\tan^2 x)+\tan^9 x(1+\tan^2 x))\dx\\ \\ =\int_{0}^{\frac{\pi}{3}}\Big(\tan^5x\Big(\dfrac{1}{\cos^2 x}-\tan^2 x+\tan ^2 x\Big) + \tan^9x\Big(\dfrac{1}{\cos^2 x}-\tan^2 x+\tan ^2 x\Big)\Big)\, dx \\ = \int_{0}^{\frac{\pi}{3}}\Big(\tan^5 x \cdot \dfrac{1}{\cos^2 x}+\tan^9 x \cdot \dfrac{1}{\cos^2 x}\Big)\, dx$

$= \int_0^{\frac{\pi}{3}}\Big(\tan^9 x+\tan^ 5 x\Big)\cdot \dfrac{1}{\cos^2 x}\, dx \\ \\ x = 0 \Rightarrow t = 0\\ x = \frac{\pi}{3} \Rightarrow t = \sqrt 3 \\ \\ I = \int_{0}^{\sqrt 3}(t^9+t^5)\, dt = \Big(\dfrac{t^{10}}{10}+\dfrac{t^6}{6}\Big)\Big|_{0}^{\sqrt 3} = \dfrac{\sqrt 3^{10}}{10}+\dfrac{\sqrt 3^6}{6} = \dfrac{3^5}{10}+\dfrac{3^3}{6}$

M-am folosit de identitatea:

$\boxed{\dfrac{1}{\cos^2 x}-\tan^2 x = 1}$