Resolva as integrais duplas
\int_{\pi / 6}^{\pi / 4} \int_{t g(x)}^{\operatorname{sec} (x)}(y+\sin (x)) d y d x

Passo 1

O enunciado pede para calcular a integral dupla
I = \int_{x=\pi/6}^{\pi/4}\int_{y=\tan x}^{\operatorname{sec} x}\bigl(y+\sin x\bigr)\,dy\,dx.
Isso envolve primeiro integrar em y (mantendo x fixo) e depois integrar o resultado em x. Vamos usar as fórmulas básicas de antiderivadas e identidades trigonométricas \operatorname{sec}^2x - \tan^2x =1 e \sin^2x =1-\cos^2x.

Passo 2

Calcular a integral interna em y:
\int_{\tan x}^{\operatorname{sec} x}(y+\sin x)\,dy = \left[\tfrac12\,y^2 + y\,\sin x\right]_{y=\tan x}^{y=\operatorname{sec} x} = \tfrac12\bigl(\operatorname{sec}^2x-\tan^2x\bigr) + \sin x\bigl(\operatorname{sec} x-\tan x\bigr).
Como \operatorname{sec}^2x-\tan^2x=1, isso fica
\tfrac12 + \sin x\,\operatorname{sec} x - \sin x\,\tan x.

Passo 3

Simplificar cada termo:
  • \sin x\,\operatorname{sec} x = \tan x,
  • \sin x\,\tan x = \dfrac{\sin^2x}{\cos x} = \dfrac{1-\cos^2x}{\cos x} = \operatorname{sec} x - \cos x.
    Logo o integrando em x é
\tfrac12 + \tan x -\bigl(\operatorname{sec} x-\cos x\bigr) = \tfrac12 + \tan x - \operatorname{sec} x + \cos x.
Portanto
I = \int_{\pi/6}^{\pi/4}\Bigl(\tfrac12 + \tan x - \operatorname{sec} x + \cos x\Bigr)\,dx.

Passo 4

Encontrar uma primitiva F(x):
\int\tfrac12\,dx = \tfrac{x}{2},\quad \int\tan x\,dx = -\ln\bigl|\cos x\bigr|,\quad \int(-\operatorname{sec} x)\,dx = -\ln\bigl|\operatorname{sec} x+\tan x\bigr|,\quad \int\cos x\,dx = \sin x.
Somando,
F(x)=\frac{x}{2}-\ln\bigl|\cos x\bigr|-\ln\bigl|\operatorname{sec} x+\tan x\bigr|+\sin x.
Usando \cos x(\operatorname{sec} x+\tan x)=1+\sin x e removendo valor absoluto,
F(x)=\frac{x}{2}+\sin x-\ln\bigl(1+\sin x\bigr).

Passo 5

Avaliar F(x) entre \pi/6 e \pi/4:
I = F\bigl(\tfrac{\pi}{4}\bigr)-F\bigl(\tfrac{\pi}{6}\bigr) = \Bigl(\tfrac{\pi}{8}+\tfrac{\sqrt2}{2}-\ln\bigl(1+\tfrac{\sqrt2}{2}\bigr)\Bigr) - \Bigl(\tfrac{\pi}{12}+\tfrac12-\ln\bigl(\tfrac32\bigr)\Bigr).
Reunindo termos:
I = \frac{\pi}{24} + \frac{\sqrt2-1}{2} + \ln\!\Bigl(\frac{3/2}{(2+\sqrt2)/2}\Bigr) = \frac{\pi}{24} + \frac{\sqrt2-1}{2} + \ln\!\Bigl(\frac{3}{2+\sqrt2}\Bigr).

Resposta

\boxed{\,\frac{\pi}{24} \;+\;\frac{\sqrt2-1}{2}\;+\;\ln\!\bigl(\tfrac{3}{2+\sqrt2}\bigr)\,}.

Exercícios relacionados