Solve the following integral using integration by parts sin(3x)cos(5x)dx Please

Question

Solve the following integral using integration by parts sin(3x)cos(5x)dx Please show all necessary steps so I can completely understand Thank You for the help in advance (I will immediately rate)

Explanation / Answer

1.Integrate the original integrand by parts: ? sin(3x)cos(5x) dx----------> Let f'(x) = cos(5x)----------> f(x) = sin(5x) / 5----------> Let g(x) = sin(3x)----------> g'(x) = 3cos(3x)----------> ? f'(x)g(x) dx = f(x)g(x) - ? f(x)g'(x) dx----------> ? sin(3x)cos(5x) dx = sin(3x)sin(5x) / 5 - ? 3sin(5x)cos(3x) / 5 dx----------> Integrate the new integrand by parts: ? 3sin(5x)cos(3x) / 5 dx----------> Let f'(x) = sin(5x) / 5----------> f(x) = -cos(5x) / 25----------> Let g(x) = 3cos(3x)----------> g'(x) = -9sin(3x)----------> ? f'(x)g(x) dx = f(x)g(x) - ? f(x)g'(x) dx----------> ? 3sin(5x)cos(3x) / 5 dx = -3cos(3x)cos(5x) / 25 - ? 9 sin(3x)cos(5x) / 25 dx----------> ? 3sin(5x)cos(3x) / 5 dx = -3cos(3x)cos(5x) / 25 - 9 ? sin(3x)cos(5x) dx / 25----------> Put it all together and collect like terms to solve for the integral:----------> ? sin(3x)cos(5x) dx = sin(3x)sin(5x) / 5 - ? 3sin(5x)cos(3x) / 5 dx----------> ? sin(3x)cos(5x) dx = sin(3x)sin(5x) / 5 - [-3cos(3x)cos(5x) / 25 - ? 9 sin(3x)cos(5x) / 25 dx]----------> ? sin(3x)cos(5x) dx = sin(3x)sin(5x) / 5 + 3cos(3x)cos(5x) / 25 + ? 9 sin(3x)cos(5x) / 25 dx----------> ? sin(3x)cos(5x) dx = sin(3x)sin(5x) / 5 + 3cos(3x)cos(5x) / 25 + ? 9 sin(3x)cos(5x) / 25 dx----------> 16 ? sin(3x)cos(5x) dx / 25 = sin(3x)sin(5x) / 5 + 3cos(3x)cos(5x) / 25----------> ? sin(3x)cos(5x) dx = [5sin(3x)sin(5x) + 3cos(3x)cos(5x)] / 16 + C I hope it is clear now, give 5 star

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