Question

Integral de cos^2 xdx

Answer 1

Calcular $\displaystyle\int\cos^2 x\,dx.$

$\bullet~~$ Forma 1. Método de integração por partes:

$\displaystyle\int\cos^2 x\,dx\\\\\\ =\int \cos x\cdot \cos x\,dx\\\\\\\\ \begin{array}{lcl} u=\cos x&~\Rightarrow~&du=-\mathrm{sen\,}x\,dx\\\\ dv=\cos x\,dx&~\Leftarrow~&v=\mathrm{sen\,}x \end{array}\\\\\\\\ \int u\,dv=uv-\int v\,du\\\\\\ \int\cos^2 x\,dx=\mathrm{sen\,}x\cos x-\int \mathrm{sen\,}x\cdot(-\mathrm{sen\,}x)\,dx\\\\\\ \int\cos^2 x\,dx=\mathrm{sen\,}x\cos x+\int \mathrm{sen^2\,}x\,dx$

$\displaystyle\int\cos^2 x\,dx=\mathrm{sen\,}x\cos x+\int (1-\cos^2 x)\,dx\\\\\\ \int\cos^2 x\,dx=\mathrm{sen\,}x\cos x+\int dx-\int\cos^2 x\,dx\\\\\\ \int\cos^2 x\,dx=\mathrm{sen\,}x\cos x+x-\int\cos^2 x\,dx\\\\\\ 2\int\cos^2 x\,dx=\mathrm{sen\,}x\cos x+x\\\\\\ \therefore~~\boxed{\begin{array}{c}\displaystyle\int \cos^2 x\,dx=\dfrac{1}{2}\,\mathrm{sen\,}x\cos x+\dfrac{1}{2}\,x+C \end{array}}$

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$\bullet~~$ Forma 2. Usando as identidades do arco duplo:

$\displaystyle\int\cos^2 x\,dx\\\\\\ =\int\!\!\left(\dfrac{1}{2}+\dfrac{1}{2}\cos 2x \right )\!dx\\\\\\ =\dfrac{1}{2}\int dx+\dfrac{1}{2}\int\cos 2x\,dx\\\\\\ =\dfrac{1}{2}\,x+\dfrac{1}{2}\cdot \left(\dfrac{1}{2}\,\mathrm{sen\,}2x \right )+C\\\\\\ =\dfrac{1}{2}\,x+\dfrac{1}{4}\,\mathrm{sen\,}2x+C\\\\\\ =\dfrac{1}{2}\,x+\dfrac{1}{4}\cdot 2\,\mathrm{sen\,}x\cos x+C\\\\\\ \therefore~~\boxed{\begin{array}{c}\displaystyle\int\cos^2 x\,dx=\dfrac{1}{2}\,x+\dfrac{1}{2}\,\mathrm{sen\,}x\cos x+C \end{array}}$