\left.\int \frac{\sqrt[3]{\tan ^{2} x} d x}{\cos ^{2} x} \quad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left\lvert\, \operatorname{resp} . \quad \frac{3}{5}(\operatorname{tg} x)^{5 / 3}+C\right.\right) \int \frac{d x}{2 \operatorname{sen}^{2} x+3 \cos ^{2} x} \quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots\left(\right. resp. \left.\frac{1}{\sqrt{6}} \operatorname{arctg}\left(\sqrt{\frac{2}{3}} \operatorname{tg} x\right)+C\right) Passo 1Temos a integral \int \frac{\sqrt[3]{\tan^{2}x}}{\cos^{2}x}\,dx. Repare que \sqrt[3]{\tan^{2}x}=(\tan x)^{2/3} e que 1/\cos^{2}x=\operatorname{sec}^{2}x. O conteúdo aqui é usar a substituição u=\tan x, pois du=\operatorname{sec}^{2}x\,dx. Passo 2Fazendo u=\tan x temos du=\operatorname{sec}^{2}x\,dx \quad\Longrightarrow\quad \frac{dx}{\cos^{2}x}=du. A integral vira \int (\tan x)^{2/3}\operatorname{sec}^{2}x\,dx =\int u^{2/3}\,du. Integrando em u, \int u^{2/3}\,du =\frac{u^{5/3}}{5/3}+C =\frac{3}{5}u^{5/3}+C. Passo 3Agora consideramos \int \frac{dx}{2\sin^{2}x+3\cos^{2}x}. O denominador pode ser reescrito como 2\sin^{2}x+3\cos^{2}x =2(1-\cos^{2}x)+3\cos^{2}x =2+\cos^{2}x. Usaremos a substituição t=\tan x, pois aí dt=\operatorname{sec}^{2}x\,dx e \cos^{2}x=1/(1+t^{2}). Passo 4Fazendo t=\tan x, dx=dt/(1+t^{2}) e 2+\cos^{2}x =2+\frac{1}{1+t^{2}} =\frac{2(1+t^{2})+1}{1+t^{2}} =\frac{3+2t^{2}}{1+t^{2}}. A integral torna-se \int \frac{dx}{2+\cos^{2}x} =\int \frac{dt/(1+t^{2})}{(3+2t^{2})/(1+t^{2})} =\int \frac{dt}{3+2t^{2}}. Passo 5Escreva \int \frac{dt}{3+2t^{2}} =\frac{1}{2}\int \frac{dt}{\tfrac{3}{2}+t^{2}}. Como \tfrac{3}{2}=(\sqrt{3/2})^{2}, temos \int \frac{dt}{t^{2}+(\sqrt{3/2})^{2}} =\frac{1}{\sqrt{3/2}}\arctan\!\bigl(t/\sqrt{3/2}\bigr)+C. Multiplicando por 1/2 dá \frac{1}{2}\cdot\frac{1}{\sqrt{3/2}} =\frac{1}{\sqrt{6}}, e t/\sqrt{3/2}=\tan x\cdot\sqrt{\tfrac{2}{3}}. Resposta \frac{1}{\sqrt{6}}\;\arctan\!\Bigl(\sqrt{\tfrac{2}{3}}\;\tan x\Bigr)+C. MOSTRAR SOLUÇÃO COMPLETAExercícios relacionados \int\left(e^{5 x}+a^{5 x}\right) d x, \quad a>0 \quad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\left(\right. resp. \left.\quad\frac{e^{5 x}}{5}+\frac{a^{5 x}}{5 \ln a}+C\right) \int \ln \left(x^{2}+1\right) d x \qquad (resp. \quad x \ln \left(1+x^{2}\right)-2 x+2 \operatorname{arctg} x+C ) \int x \arctan \left(\sqrt{x^{2}-1}\right) d x \qquad \left(\right. resp. \left.\quad \frac{1}{2}\left[x^{2} \operatorname{arctg} \sqrt{x^{2}-1}-\sqrt{x^{2}-1}\right]+C\right) \int x \arctan \left(\sqrt{x^{2}-1}\right) d x \qquad \left(\right. resp. \left.\quad \frac{1}{2}\left[x^{2} \operatorname{arctg} \sqrt{x^{2}-1}-\sqrt{x^{2}-1}\right]+C\right)