Questão 8) Encontre uma formula para cada uma das seguintes integrais:
\int \operatorname{sen}^{n}(\alpha x) \cdot \cos ^{m}(\beta x) d x

Passo 1

O enunciado pede uma fórmula geral para
\int \sin^{n}(\alpha x)\,\cos^{m}(\beta x)\,dx,
onde n e m são inteiros não negativos e \alpha,\beta constantes.
Vamos usar a representação exponencial de \sin e \cos para expandir o produto em soma de termos exponenciais e, em seguida, integrar termo a termo.

Passo 2

Lembrando que
\sin(\alpha x)=\frac{e^{i\alpha x}-e^{-i\alpha x}}{2i}, \qquad \cos(\beta x)=\frac{e^{i\beta x}+e^{-i\beta x}}{2},
temos
\sin^{n}(\alpha x) =\Bigl(\frac{e^{i\alpha x}-e^{-i\alpha x}}{2i}\Bigr)^{n} =\frac{1}{(2i)^{n}} \sum_{j=0}^{n}(-1)^{j}\binom{n}{j}\,e^{i(n-2j)\alpha x},
\cos^{m}(\beta x) =\Bigl(\frac{e^{i\beta x}+e^{-i\beta x}}{2}\Bigr)^{m} =\frac{1}{2^{m}} \sum_{k=0}^{m}\binom{m}{k}\,e^{i(m-2k)\beta x}.

Passo 3

Multiplicando as duas expansões, obtemos
\sin^{n}(\alpha x)\,\cos^{m}(\beta x) =\frac{1}{2^{n+m}i^{n}} \sum_{j=0}^{n}\sum_{k=0}^{m}(-1)^{j}\binom{n}{j}\binom{m}{k} \,e^{\,i\bigl((n-2j)\alpha+(m-2k)\beta\bigr)x}.
Assim
\int \sin^{n}(\alpha x)\cos^{m}(\beta x)\,dx =\frac{1}{2^{n+m}i^{n}} \sum_{j=0}^{n}\sum_{k=0}^{m}(-1)^{j}\binom{n}{j}\binom{m}{k} \int e^{\,i\bigl((n-2j)\alpha+(m-2k)\beta\bigr)x}\,dx.

Passo 4

Cada integral exponencial é trivial:
\int e^{i\lambda x}\,dx =\frac{e^{i\lambda x}}{i\lambda}, \quad \lambda\neq 0.
Logo, colocando \lambda_{jk}=(n-2j)\alpha+(m-2k)\beta,
\int \sin^{n}(\alpha x)\cos^{m}(\beta x)\,dx =\frac{1}{2^{n+m}i^{n}} \sum_{j=0}^{n}\sum_{k=0}^{m}(-1)^{j}\binom{n}{j}\binom{m}{k} \frac{e^{\,i\lambda_{jk}x}}{i\,\lambda_{jk}} + C.

Passo 5

Reagrupando o fator 1/i^{n+1} e escrevendo a soma completa, concluímos a fórmula geral:
\boxed{ \int \sin^{n}(\alpha x)\cos^{m}(\beta x)\,dx =\frac{1}{2^{n+m}i^{n+1}} \sum_{j=0}^{n}\sum_{k=0}^{m}(-1)^{j}\binom{n}{j}\binom{m}{k} \frac{e^{\,i\bigl((n-2j)\alpha+(m-2k)\beta\bigr)x}}{(n-2j)\alpha+(m-2k)\beta} + C. }

Resposta

\int \sin^{n}(\alpha x)\cos^{m}(\beta x)\,dx =\frac{1}{2^{n+m}i^{n+1}} \sum_{j=0}^{n}\sum_{k=0}^{m}(-1)^{j}\binom{n}{j}\binom{m}{k} \frac{e^{\,i\bigl((n-2j)\alpha+(m-2k)\beta\bigr)x}}{(n-2j)\alpha+(m-2k)\beta} + C.

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