Trig-e

Calculus Level 3

∫ cos ⁡ ( 2 x ) sin ⁡ ( 5 x ) d x = ? \large \int\cos(2x) \sin(5x) \, dx= \, ?

Assume we ignore the arbitrary constant of integration .

1 2 sin ⁡ ( 3 x ) + 1 2 sin ⁡ ( 7 x ) \frac{1}{2} \, \sin(3x) + \frac{1}{2} \, \sin(7x) 1 10 ( sin ⁡ ( 2 x ) cos ⁡ ( 5 x ) − cos ⁡ ( 2 x ) sin ⁡ ( 5 x ) ) \frac{1}{10} ( \sin(2x) \, \cos(5x) - \cos(2x) \, \sin(5x) ) − 1 6 cos ⁡ ( 3 x ) − 1 14 cos ⁡ ( 7 x ) -\frac{1}{6} \, \cos(3x) - \frac{1}{14} \, \cos(7x) − 1 2 sin ⁡ ( 2 x ) − 1 5 sin ⁡ ( 5 x ) -\frac{1}{2} \, \sin(2x) - \frac{1}{5} \, \sin(5x)

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Brandon Stocks by Brandon Stocks , 45, USA

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2 solutions

Brandon Stocks
Jul 1, 2016

c o s ( θ ) = e i θ + e − i θ 2 a n d s i n ( θ ) = e i θ − e − i θ 2 i \large cos(\theta) = \frac{ e^{i\theta} + e^{-i\theta} }{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; and \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; sin(\theta) = \frac{ e^{i\theta} - e^{-i\theta} }{2i}

c o s ( 2 x ) s i n ( 5 x ) = ( e i 2 x + e − i 2 x ) ( e i 5 x − e − i 5 x ) 4 i cos(2x) \, sin(5x) = \frac{ (e^{i2x} + e^{-i2x})(e^{i5x} - e^{-i5x}) }{4i}

Recall that − i = 1 / i -i = 1/i

c o s ( 2 x ) s i n ( 5 x ) = i 4 ( e i 2 x + e − i 2 x ) ( e − i 5 x − e i 5 x ) = i 4 ( e − i 3 x − e i 7 x + e − i 7 x − e i 3 x ) cos(2x) \, sin(5x) = \frac{i}{4} \, (e^{i2x} + e^{-i2x})(e^{-i5x} - e^{i5x}) = \frac{i}{4} \, (e^{-i3x} - e^{i7x} + e^{-i7x} - e^{i3x} )

Applying Euler's Formula

e − i 3 x − e i 3 x = c o s ( − 3 x ) + i s i n ( − 3 x ) − ( c o s ( 3 x ) + i s i n ( 3 x ) ) = − 2 i s i n ( 3 x ) e^{-i3x} - e^{i3x} = cos(-3x) + i \, sin(-3x) - \Big( cos(3x) + i \, sin(3x) \Big) = -2i \, sin(3x)

Likewise e − i 7 x − e i 7 x = − 2 i s i n ( 7 x ) \; \; \; e^{-i7x} - e^{i7x} = -2i \, sin(7x)

c o s ( 2 x ) s i n ( 5 x ) = i 4 ( − 2 i s i n ( 3 x ) − 2 i s i n ( 7 x ) ) = 1 2 ( s i n ( 3 x ) + s i n ( 7 x ) ) cos(2x) \, sin(5x) = \frac{i}{4} \, \Big( -2i \, sin(3x) - 2i \, sin(7x) \Big) = \frac{1}{2} \, \Big( sin(3x) + sin(7x) \Big)

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u =3x, du = 3 dx

∫ s i n ( 3 x ) d x = 1 3 ∫ s i n ( u ) d u = − 1 3 c o s ( u ) + C = − 1 3 c o s ( 3 x ) \int sin(3x) \, dx = \frac{1}{3} \, \int sin(u) \, du = -\frac{1}{3} \, cos(u) + C = -\frac{1}{3} \, cos(3x)

∫ s i n ( 7 x ) d x = − 1 7 c o s ( 7 x ) \int sin(7x) \, dx = -\frac{1}{7} \, cos(7x)

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∫ c o s ( 2 x ) s i n ( 5 x ) d x = − 1 6 c o s ( 3 x ) − 1 14 c o s ( 7 x ) \int cos(2x) \, sin(5x) \, dx = -\frac{1}{6} \, cos(3x) - \frac{1}{14} \, cos(7x)

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Or you can linearize the integrand by applying the Prosthaphaeresis Formulas , to get cos ⁡ ( 2 x ) sin ⁡ ( 5 x ) = 1 2 ( sin ⁡ ( 7 x ) + sin ⁡ ( 3 x ) ) \cos(2x) \sin(5x)= \dfrac12 ( \sin(7x) + \sin(3x) ) .

Pi Han Goh - 4 years, 11 months ago

Yes but this is a good way to learn about the exponential form of trig functions. Besides, I can never remember those formulas.

Brandon Stocks - 4 years, 11 months ago
Prakhar Bindal
Jul 2, 2016

Using 2sinAcosB = sin(A+B)+Sin(A-B) will yield answer in a line!

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