Calcule as integrais indefinidas e integrais definidas.
\int \operatorname{sen}^{2}(\pi t) \cos ^{5}(\pi t) d t

Passo 1

O enunciado pede para calcular a integral indefinida
\int \operatorname{sen}^{2}(\pi t)\,\cos^{5}(\pi t)\,dt.
O conteúdo é integração de função trigonométrica usando identidade de potência e substituição.

Passo 2

Observamos que \cos^{5}(\pi t)=\cos^{4}(\pi t)\,\cos(\pi t) e, usando \cos^{2}x=1-\operatorname{sen}^{2}x, temos
\cos^{4}(\pi t)=\bigl(\cos^{2}(\pi t)\bigr)^{2}=\bigl(1-\operatorname{sen}^{2}(\pi t)\bigr)^{2}.
Assim a integral vira
\int \operatorname{sen}^{2}(\pi t)\bigl(1-\operatorname{sen}^{2}(\pi t)\bigr)^{2}\cos(\pi t)\,dt.

Passo 3

Fazemos a substituição
u=\operatorname{sen}(\pi t) \quad\Longrightarrow\quad du=\pi\,\cos(\pi t)\,dt \quad\Longrightarrow\quad \cos(\pi t)\,dt=\frac{du}{\pi}.
Então
\int \operatorname{sen}^{2}(\pi t)\bigl(1-\operatorname{sen}^{2}(\pi t)\bigr)^{2}\cos(\pi t)\,dt =\int u^{2}(1-u^{2})^{2}\,\frac{du}{\pi} =\frac{1}{\pi}\int u^{2}(1-u^{2})^{2}\,du.

Passo 4

Expandimos o integrando em u:
u^{2}(1-u^{2})^{2}=u^{2}(1-2u^{2}+u^{4}) =u^{2}-2u^{4}+u^{6}.
Logo
\frac{1}{\pi}\int\bigl(u^{2}-2u^{4}+u^{6}\bigr)\,du =\frac{1}{\pi}\Bigl(\frac{u^{3}}{3}-2\frac{u^{5}}{5}+\frac{u^{7}}{7}\Bigr)+C.

Passo 5

Voltando a variável t com u=\operatorname{sen}(\pi t), obtemos
\int \operatorname{sen}^{2}(\pi t)\cos^{5}(\pi t)\,dt =\frac{1}{\pi}\Bigl(\frac{\operatorname{sen}^{3}(\pi t)}{3}-\frac{2\,\operatorname{sen}^{5}(\pi t)}{5}+\frac{\operatorname{sen}^{7}(\pi t)}{7}\Bigr)+C.

Resposta

\int \operatorname{sen}^{2}(\pi t)\cos^{5}(\pi t)\,dt =\frac{1}{\pi}\Bigl(\frac{\operatorname{sen}^{3}(\pi t)}{3}-\frac{2\,\operatorname{sen}^{5}(\pi t)}{5}+\frac{\operatorname{sen}^{7}(\pi t)}{7}\Bigr)+C.

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