Calcule.
\int_{0}^{\pi / 4} \operatorname{sec} ^{4}(x) \operatorname{tg}^{4}(x) d x

Passo 1

O enunciado pede para calcular a integral
\int_{0}^{\pi/4} \operatorname{sec}^{4}(x)\,\tan^{4}(x)\,dx.
O conteúdo usado será a identidade \tan^{2}x = \operatorname{sec}^{2}x - 1 para expandir a integral em potências de \operatorname{sec} x e depois a substituição u = \tan x, com du = \operatorname{sec}^{2}x\,dx, para simplificar o cálculo.

Passo 2

Usando \tan^{4}x = (\operatorname{sec}^{2}x - 1)^{2}, temos
\operatorname{sec}^{4}x\,\tan^{4}x = \operatorname{sec}^{4}x\,(\operatorname{sec}^{2}x - 1)^{2} = \operatorname{sec}^{4}x\,\bigl(\operatorname{sec}^{4}x - 2\operatorname{sec}^{2}x + 1\bigr) = \operatorname{sec}^{8}x \;-\;2\,\operatorname{sec}^{6}x \;+\;\operatorname{sec}^{4}x.
Logo
\int_{0}^{\pi/4}\operatorname{sec}^{4}x\,\tan^{4}x\,dx = \int_{0}^{\pi/4}\bigl(\operatorname{sec}^{8}x - 2\operatorname{sec}^{6}x + \operatorname{sec}^{4}x\bigr)\,dx.

Passo 3

Fazendo u = \tan x, du = \operatorname{sec}^{2}x\,dx. Observe:
\operatorname{sec}^{8}x\,dx = \operatorname{sec}^{6}x\,\operatorname{sec}^{2}x\,dx = (1+u^{2})^{3}\,du,
\operatorname{sec}^{6}x\,dx = \operatorname{sec}^{4}x\,\operatorname{sec}^{2}x\,dx = (1+u^{2})^{2}\,du,
\operatorname{sec}^{4}x\,dx = \operatorname{sec}^{2}x\,\operatorname{sec}^{2}x\,dx = (1+u^{2})\,du.
Portanto os antiderivados são:
\int\operatorname{sec}^{8}x\,dx = \int(1+u^{2})^{3}du = u + u^{3} + \tfrac{3}{5}u^{5} + \tfrac{1}{7}u^{7},
\int\operatorname{sec}^{6}x\,dx = \int(1+u^{2})^{2}du = u + \tfrac{2}{3}u^{3} + \tfrac{1}{5}u^{5},
\int\operatorname{sec}^{4}x\,dx = \int(1+u^{2})du = u + \tfrac{1}{3}u^{3}.

Passo 4

Voltando à integral original e avaliando de x=0 a x=\pi/4 (isto é, de u=0 a u=1), obtemos
\Bigl[u + u^{3} + \tfrac{3}{5}u^{5} + \tfrac{1}{7}u^{7}\Bigr]_{0}^{1} \;-\;2\Bigl[u + \tfrac{2}{3}u^{3} + \tfrac{1}{5}u^{5}\Bigr]_{0}^{1} \;+\;\Bigl[u + \tfrac{1}{3}u^{3}\Bigr]_{0}^{1}.
Em u=1 isso vale
\Bigl(1+1+\tfrac{3}{5}+\tfrac{1}{7}\Bigr) -2\Bigl(1+\tfrac{2}{3}+\tfrac{1}{5}\Bigr) +\Bigl(1+\tfrac{1}{3}\Bigr) = \frac{96}{35}-\frac{56}{15}+\frac{4}{3} = \frac{12}{35}.

Resposta

\boxed{\frac{12}{35}}.

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