definite integral

por **Guill** » Ter Jul 26, 2011 23:01

$\int_{0}^{\frac{\pi}{2}}cos^2x.ln\left(cos x \right)dx$

Podemos usar a integração por partes:

u = ln(cos x)
du = $\frac{1}{cosx}$ dx

v = $\frac{x+sen2x}{4}$
dv = cos²x dx

Substituindo:

$ln(cosx).\frac{x+sen2x}{4}-\int_{0}^{\frac{\pi}{2}}\frac{x+sen2x}{4}.\frac{1}{cosx}dx$

$ln(cosx).\frac{x+sen2x}{4}-\frac{1}{4}.\int_{0}^{\frac{\pi}{2}}\frac{x}{cosx}+\frac{sen2x}{cosx}$

$ln(cosx).\frac{x+sen2x}{4}-\frac{1}{4}.\int_{0}^{\frac{\pi}{2}}\frac{x}{cosx}+\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\frac{sen2x}{cosx}$

Resolvendo as integrais:

$ln(cosx).\frac{x+sen2x}{4}-\frac{1}{4}.\int_{0}^{\frac{\pi}{2}}\frac{x}{cosx}-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\frac{sen2x}{cosx}$

$ln(cosx).\frac{x+sen2x}{4}-\frac{1}{4}.\left(\frac{x^2}{ln(cosx)}-\int_{0}^{\frac{\pi}{2}}ln(cosx)dx \right)-\frac{1}{2}.\left(\int_{0}^{\frac{\pi}{2}}senx dx \right)$

$ln(cosx).\frac{x+sen2x}{4}-\frac{x^2}{4.ln(cosx)}+\frac{1}{2.cosx}}-\frac{cosx}{2} + C$

por **VtinxD** » Qua Jul 27, 2011 22:18

Corrigindo: du= tg(x)dx (por causa da regra da cadeia)

definite integral

definite integral

definite integral

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