Ensino Superior ⇒ Integrais indefinidas

[emanuel9393MOD]

Determine:
[tex3]\int \tan^3 x \cdot \cos x \ dx[/tex3]

Resposta

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Gabarito: [tex3]\sec x + \cos x + k[/tex3]

Grande abraço!

[NiltonGMJr]

[tex3]\int\limits_{}^{}\tan ^3x \cdot \cos x \cdot dx=\int\limits_{}^{}\frac{\tan ^3x}{\sec x } \cdot dx[/tex3]

Fazendo [tex3]u=\sec (x),\, du=\sec (x)\tan (x)dx[/tex3], Logo [tex3]dx=\frac{du}{\sec (x)\tan (x)}[/tex3]

[tex3]\int\limits_{}^{}\frac{\tan ^3x}{\sec ^2x \cdot \tan x } \cdot du=\int\limits_{}^{}\frac{\tan ^2x}{\sec ^2x} \cdot du=\int\limits_{}^{}\frac{\tan ^2x}{u^2} \cdot du[/tex3]

Lembrando que: [tex3]\sec^2(x)=1+\tan^2(x)[/tex3], temos que [tex3]\tan^2(x)=\sec^2(x)-1[/tex3]

[tex3]\int\limits_{}^{}\frac{\sec ^2x-1}{u^2} du=\int\limits_{}^{}\frac{u^2-1}{u^2} du=\int\limits_{}^{}\ du-\int\limits_{}^{}\frac{1}{u^2} du=u+\frac{1}{u}+k=\sec x +\frac{1}{\sec x }+k=\sec x +\cos x +k[/tex3]

[emanuel9393MOD]

Obrigado!

[NiltonGMJr]

Sem problemas, disponha sempre.