Calcule.
\int\left(\frac{1}{2}+\frac{1}{2} \cos 2 x\right)^{2} d x

Passo 1

O enunciado pede para calcular
\int\Bigl(\tfrac12 + \tfrac12\cos2x\Bigr)^{2}\,dx.
Aqui vamos usar a expansão de quadrado, identidade de redução de potência para \cos^{2} e depois integrar termo a termo.

Passo 2

Primeiro, expandimos o quadrado usando (a+b)^{2}=a^{2}+2ab+b^{2}. Com a=\tfrac12 e b=\tfrac12\cos2x temos
\Bigl(\tfrac12 + \tfrac12\cos2x\Bigr)^{2} = \tfrac14 + 2\cdot\tfrac12\cdot\tfrac12\cos2x + \tfrac14\cos^{2}2x = \tfrac14 + \tfrac12\cos2x + \tfrac14\cos^{2}2x.

Passo 3

Aplicamos a identidade de redução de potência
\cos^{2}u = \tfrac12\bigl(1+\cos2u\bigr),
com u=2x. Logo
\cos^{2}2x = \tfrac12\bigl(1+\cos4x\bigr).

Passo 4

Substituindo em nosso integrando,
\begin{aligned} \tfrac14 + \tfrac12\cos2x + \tfrac14\cos^{2}2x &= \tfrac14 + \tfrac12\cos2x + \tfrac14\cdot\tfrac12\bigl(1+\cos4x\bigr)\\\ &= \tfrac14 + \tfrac12\cos2x + \tfrac18 + \tfrac18\cos4x\\\ &= \tfrac38 + \tfrac12\cos2x + \tfrac18\cos4x. \end{aligned}
Assim resta integrar
\int\Bigl(\tfrac38 + \tfrac12\cos2x + \tfrac18\cos4x\Bigr)dx.

Passo 5

Integramos termo a termo:
\int \tfrac38\,dx = \tfrac38\,x,
\int \tfrac12\cos2x\,dx = \tfrac12\cdot\tfrac12\sin2x = \tfrac14\sin2x,
\int \tfrac18\cos4x\,dx = \tfrac18\cdot\tfrac14\sin4x = \tfrac1{32}\sin4x.

Resposta

\int\Bigl(\tfrac12 + \tfrac12\cos2x\Bigr)^{2}dx = \tfrac38\,x + \tfrac14\sin2x + \tfrac1{32}\sin4x + C.

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