Integrais trigonometricas: \int \sin ^{m} x \cos ^{n} x d x, m e n par
\cos ^{2} x =\frac{1}{2}(1+\cos 2 x)
\sin ^{2} x =\frac{1}{2}(1-\cos 2 x)
4. \int \sin ^{2} x \cos ^{6} x d x

Passo 1

O enunciado pede para calcular a integral
\int \sin^{2}x\;\cos^{6}x\;dx
usando técnicas de integrais trigonométricas quando os expoentes são pares. Vamos aplicar as identidades de ângulo duplo
\sin^{2}x = \tfrac12\bigl(1-\cos2x\bigr), \quad \cos^{2}x = \tfrac12\bigl(1+\cos2x\bigr),
e depois expandir e integrar termo a termo.

Passo 2

Primeiro escrevemos \cos^{6}x como (\cos^{2}x)^{3} e substituímos as identidades:
\sin^{2}x\,\cos^{6}x = \Bigl(\tfrac12(1-\cos2x)\Bigr)\,\Bigl(\tfrac12(1+\cos2x)\Bigr)^{3}.
Colocando o fator comum \tfrac1{2}\cdot\tfrac1{8}=\tfrac1{16},
\sin^{2}x\,\cos^{6}x = \tfrac1{16}\,(1-\cos2x)\,(1+\cos2x)^{3}.

Passo 3

Agora expandimos (1-\cos2x)(1+\cos2x)^{3}. Chamando c=\cos2x,
(1-c)(1+c)^{3} =(1-c)(1+3c+3c^2+c^3) =1+2c-2c^3-c^4.
Logo
\sin^{2}x\,\cos^{6}x =\tfrac1{16}\bigl(1 + 2\cos2x -2\cos^{3}2x - \cos^{4}2x\bigr).

Passo 4

Integramos termo a termo:
  1. \displaystyle\int 1\,dx = x.
  2. \displaystyle\int 2\cos2x\,dx = \sin2x.
  3. Para \cos^{3}2x usamos
\cos^{3}2x = \tfrac14\bigl(3\cos2x + \cos6x\bigr),
então
\int \cos^{3}2x\,dx =\tfrac14\Bigl(3\int\cos2x\,dx + \int\cos6x\,dx\Bigr) =\tfrac14\Bigl(3\cdot\tfrac12\sin2x + \tfrac16\sin6x\Bigr) =\tfrac38\sin2x + \tfrac1{24}\sin6x.
  1. Para \cos^{4}2x usamos
\cos^{4}2x = \tfrac18\bigl(3 + 4\cos4x + \cos8x\bigr),
logo
\int \cos^{4}2x\,dx =\tfrac18\Bigl(3x + 4\cdot\tfrac14\sin4x + \tfrac18\sin8x\Bigr) =\tfrac38\,x + \tfrac18\sin4x + \tfrac1{64}\sin8x.

Passo 5

Juntando tudo com o fator \tfrac1{16}:
\int \sin^{2}x\,\cos^{6}x\,dx =\tfrac1{16}\Bigl[ x + \sin2x -2\bigl(\tfrac38\sin2x + \tfrac1{24}\sin6x\bigr) -\bigl(\tfrac38x + \tfrac18\sin4x + \tfrac1{64}\sin8x\bigr) \Bigr]+C.
Simplificando dentro dos colchetes:
x - \tfrac38x = \tfrac58x, \quad \sin2x -2\cdot\tfrac38\sin2x = \tfrac14\sin2x,
-2\cdot\tfrac1{24}\sin6x = -\tfrac1{12}\sin6x, \quad -\,\tfrac18\sin4x,\quad -\,\tfrac1{64}\sin8x.
Então
\int \sin^{2}x\,\cos^{6}x\,dx =\tfrac1{16}\Bigl(\tfrac58x + \tfrac14\sin2x - \tfrac18\sin4x - \tfrac1{12}\sin6x - \tfrac1{64}\sin8x\Bigr)+C.

Resposta

\boxed{ \int \sin^{2}x\,\cos^{6}x\,dx = \frac{5x}{128} + \frac{\sin2x}{64} - \frac{\sin4x}{128} - \frac{\sin6x}{192} - \frac{\sin8x}{1024} + C }.

Exercícios relacionados