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Calcule as seguintes integrais:
  • \int \sin 4 x \cos 5 x d x;
  • \int \sin 8 x \cos 5 x d x;
  • \int \cos \pi x \cos 4 \pi x d x;
  • \int \sin 5 \theta \sin \theta d \theta;
  • \int_{0}^{\pi / 2} \operatorname{sen} 2 x \cos 3 x d x.

Passo 1

O enunciado pede várias integrais de produto de funções trigonométricas, incluindo uma integral definida. Para resolver, vamos usar as identidades produto-para-soma:
\sin A \cos B = \frac{1}{2}\bigl[\sin(A+B)+\sin(A-B)\bigr],
\cos A \cos B = \frac{1}{2}\bigl[\cos(A-B)+\cos(A+B)\bigr],
\sin A \sin B = \frac{1}{2}\bigl[\cos(A-B)-\cos(A+B)\bigr].

Passo 2

Para \displaystyle \int \sin 4x \cos 5x \,dx, aplicamos
\sin 4x \cos 5x = \tfrac12\bigl[\sin(4x+5x)+\sin(4x-5x)\bigr] = \tfrac12\bigl[\sin(9x)+\sin(-x)\bigr] = \tfrac12\bigl[\sin(9x)-\sin x\bigr].
Logo
\int \sin 4x \cos 5x \,dx = \tfrac12\int\bigl[\sin(9x)-\sin x\bigr]\,dx = \tfrac12\Bigl(-\tfrac{1}{9}\cos 9x + \cos x\Bigr)+C = -\tfrac{1}{18}\cos 9x + \tfrac12\cos x + C.

Passo 3

Para \displaystyle \int \sin 8x \cos 5x \,dx, fazemos
\sin 8x \cos 5x = \tfrac12\bigl[\sin(8x+5x)+\sin(8x-5x)\bigr] = \tfrac12\bigl[\sin 13x + \sin 3x\bigr].
Então
\int \sin 8x \cos 5x \,dx = \tfrac12\int\bigl[\sin 13x + \sin 3x\bigr]\,dx = \tfrac12\Bigl(-\tfrac{1}{13}\cos 13x - \tfrac{1}{3}\cos 3x\Bigr)+C = -\tfrac{1}{26}\cos 13x - \tfrac{1}{6}\cos 3x + C.

Passo 4

Para \displaystyle \int \cos(\pi x)\cos(4\pi x)\,dx, usamos
\cos A\cos B = \tfrac12\bigl[\cos(A-B)+\cos(A+B)\bigr],
com A=\pi x, B=4\pi x:
\cos(\pi x)\cos(4\pi x) = \tfrac12\bigl[\cos(-3\pi x)+\cos(5\pi x)\bigr] = \tfrac12\bigl[\cos(3\pi x)+\cos(5\pi x)\bigr].
Integramos:
\int \cos(\pi x)\cos(4\pi x)\,dx = \tfrac12\Bigl(\int\cos(3\pi x)\,dx + \int\cos(5\pi x)\,dx\Bigr) = \tfrac12\Bigl(\tfrac{1}{3\pi}\sin(3\pi x) + \tfrac{1}{5\pi}\sin(5\pi x)\Bigr)+C = \tfrac{1}{6\pi}\sin(3\pi x) + \tfrac{1}{10\pi}\sin(5\pi x) + C.

Passo 5

Para \displaystyle \int \sin 5\theta \sin \theta \,d\theta, usamos
\sin A\sin B = \tfrac12\bigl[\cos(A-B)-\cos(A+B)\bigr],
com A=5\theta, B=\theta:
\sin 5\theta \sin \theta = \tfrac12\bigl[\cos(4\theta)-\cos(6\theta)\bigr].
Logo
\int \sin 5\theta \sin \theta \,d\theta = \tfrac12\int\bigl[\cos 4\theta - \cos 6\theta\bigr]\,d\theta = \tfrac12\Bigl(\tfrac{1}{4}\sin 4\theta - \tfrac{1}{6}\sin 6\theta\Bigr)+C = \tfrac{1}{8}\sin 4\theta - \tfrac{1}{12}\sin 6\theta + C.

Passo 6

Para \displaystyle \int_{0}^{\pi/2}\sin 2x \cos 3x\,dx, voltamos à identidade
\sin 2x\cos 3x = \tfrac12\bigl[\sin(5x)+\sin(-x)\bigr] = \tfrac12\bigl[\sin 5x - \sin x\bigr].
Então
\int_{0}^{\pi/2}\sin 2x\cos 3x\,dx = \tfrac12\int_{0}^{\pi/2}\bigl[\sin 5x - \sin x\bigr]\,dx = \tfrac12\Bigl[-\tfrac{1}{5}\cos 5x + \cos x\Bigr]_{0}^{\pi/2}.
Em x=\pi/2 vem -\frac{1}{5}\cos\bigl(\tfrac{5\pi}{2}\bigr)+\cos\bigl(\tfrac{\pi}{2}\bigr)=0.
Em x=0 vem -\tfrac{1}{5}\cos 0 + \cos 0 = -\tfrac{1}{5}+1 = \tfrac{4}{5}.
A diferença é 0 - \tfrac{4}{5} = -\tfrac{4}{5}, logo o valor da integral é
\tfrac12\Bigl(-\tfrac{4}{5}\Bigr) = -\tfrac{2}{5}.

Resposta

(a) -\tfrac{1}{18}\cos 9x + \tfrac12\cos x + C
(b) -\tfrac{1}{26}\cos 13x - \tfrac{1}{6}\cos 3x + C
(c) \tfrac{1}{6\pi}\sin(3\pi x) + \tfrac{1}{10\pi}\sin(5\pi x) + C
(d) \tfrac{1}{8}\sin 4\theta - \tfrac{1}{12}\sin 6\theta + C
(e) -\tfrac{2}{5}

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