Half-angle Identities?

Calculus Level 5

∫ 0 π 2 sin ⁡ ( 3 x ) sin ⁡ ( 5 x 2 ) sin ⁡ ( x 2 ) d x \large \int_{0}^{\frac{\pi}{2}}\dfrac{\sin\left(3x\right)\sin\left(\frac{5x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \, dx

If the above integral can be expressed as a b \dfrac{a}{b} , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 53.

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Couper Leo by Couper Leo , 21, USA

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2 solutions

Kushal Bose
Nov 30, 2016

I = ∫ 0 π 2 sin ⁡ 3 x sin ⁡ 5 x / 2 sin ⁡ x / 2 d x I=\int_{0}^{\frac{\pi}{2}} \dfrac{\sin 3x \sin 5x/2 }{\sin x/2} dx

= ∫ 0 π 2 sin ⁡ 3 x sin ⁡ ( 3 x − x / 2 ) sin ⁡ x / 2 d x =\int_{0}^{\frac{\pi}{2}} \dfrac{\sin 3x \sin (3x-x/2)}{\sin x/2} dx

= ∫ 0 π 2 sin ⁡ 3 x ( sin ⁡ 3 x cos ⁡ x / 2 − cos ⁡ 3 x sin ⁡ x / 2 ) sin ⁡ x / 2 d x =\int_{0}^{\frac{\pi}{2}} \dfrac{\sin 3x(\sin 3x \cos x/2 - \cos 3x \sin x/2)}{\sin x/2} dx

= ∫ 0 π 2 s i n 2 ( 3 x ) cos ⁡ x sin ⁡ x / 2 d x − ∫ 0 π 2 sin ⁡ ( 3 x ) cos ⁡ ( 3 x ) d x =\int_{0}^{\frac{\pi}{2}} \dfrac{sin^2(3x) \cos x}{\sin x/2} dx - \int_{0}^{\frac{\pi}{2}} \sin(3x)\cos(3x) dx

For the second part 1 / 2 × ∫ 0 π 2 sin ⁡ ( 6 x ) d x = 1 / 6 1/2 \times \int_{0}^{\frac{\pi}{2}} \sin (6x) dx=1/6

Now for the first part: :

∫ 0 π 2 sin ⁡ 2 ( 3 x ) cos ⁡ x / 2 sin ⁡ x / 2 sin ⁡ x / 2 sin ⁡ x / 2 d x \int_{0}^{\frac{\pi}{2}} \dfrac{\sin^2(3x) \cos x/2 \sin x/2}{\sin x/2 \sin x/2} dx

= ∫ 0 π 2 sin ⁡ 2 ( 3 x ) sin ⁡ x sin ⁡ 2 ( x / 2 ) d x = \int_{0}^{\frac{\pi}{2}} \dfrac{\sin^2(3x) \sin x}{\sin^2(x/2)} dx

= ∫ 0 π 2 sin ⁡ 2 ( 3 x ) sin ⁡ x 1 − cos ⁡ x d x = \int_{0}^{\frac{\pi}{2}} \dfrac{\sin^2(3x) \sin x}{1-\cos x} dx

= ∫ 0 π 2 sin ⁡ 2 ( 3 x ) sin ⁡ x ( 1 + cos ⁡ x ) ( 1 − cos ⁡ x ) ( 1 + cos ⁡ x ) d x = \int_{0}^{\frac{\pi}{2}} \dfrac{\sin^2(3x) \sin x(1+\cos x)}{(1-\cos x)(1+ \cos x)} dx

= ∫ 0 π 2 ( 3 sin ⁡ ( x ) − 4 sin ⁡ 3 ( x ) ) 2 sin ⁡ x ( 1 + cos ⁡ x ) sin ⁡ 2 ( x ) d x =\int_{0}^{\frac{\pi}{2}} \dfrac{(3 \sin (x)-4\sin^3(x))^2 \sin x (1+\cos x)}{\sin^2(x)} dx

= ∫ 0 π 2 ( 3 − 4 sin ⁡ 2 x ) 2 sin ⁡ x ( 1 + cos ⁡ x ) d x =\int_{0}^{\frac{\pi}{2}} (3-4\sin^2x)^2 \sin x (1+\cos x) dx

= ∫ 0 π 2 ( 3 sin ⁡ ( x − 4 ) sin ⁡ 3 ( x ) ) ( 3 − 4 sin ⁡ 2 x ) ( 1 + cos ⁡ x ) d x =\int_{0}^{\frac{\pi}{2}} (3\sin (x-4)\sin^3(x))(3-4\sin^2x) (1+\cos x) dx

= ∫ 0 π 2 sin ⁡ 3 x ( 3 − 2 ( 1 − cos ⁡ ( 2 x ) ) ) ( 1 + cos ⁡ x ) d x =\int_{0}^{\frac{\pi}{2}} \sin 3x(3-2(1-\cos(2x)))(1+\cos x) dx

= ∫ 0 π 2 sin ⁡ 3 x ( 2 cos ⁡ ( 2 x ) + 1 ) ( 1 + cos ⁡ x ) d x =\int_{0}^{\frac{\pi}{2}} \sin3x (2 \cos (2x)+1)(1+\cos x) dx

= ∫ 0 π 2 sin ⁡ 3 x ( 2 cos ⁡ ( 2 x ) cos ⁡ ( x + 2 ) cos ⁡ ( 2 x ) + cos ⁡ x + 1 ) d x =\int_{0}^{\frac{\pi}{2}} \sin3x (2 \cos (2 x) \cos (x +2) \cos (2 x)+ \cos x +1) dx

Using product-sum rules of trigonometry :

= ∫ 0 π 2 sin ⁡ 3 x ( cos ⁡ 3 x + 2 cos ⁡ 2 x + 2 cos ⁡ x + 1 ) d x =\int_{0}^{\frac{\pi}{2}} \sin 3 x(\cos 3x + 2 \cos 2x+ 2 \cos x+1) dx

= ∫ 0 π 2 ( sin ⁡ ( 3 x ) cos ⁡ ( 3 x ) + 2 sin ⁡ ( 3 x ) cos ⁡ ( x ) + 2 sin ⁡ ( 3 x ) cos ⁡ ( 2 x ) + sin ⁡ ( 3 x ) ) d x =\int_{0}^{\frac{\pi}{2}} (\sin(3x) \cos(3x) + 2 \sin(3x)\cos(x) + 2 \sin(3x) \cos(2x) + \sin(3x)) dx

Again using product-sum rules of trigonometry :

= ∫ 0 π 2 ( sin ⁡ ( 6 x ) 2 + ( sin ⁡ ( 4 x ) + sin ⁡ ( 2 x ) ) + ( sin ⁡ ( 5 x ) + sin ⁡ ( x ) ) + sin ⁡ ( 3 x ) ) d x =\int_{0}^{\frac{\pi}{2}} (\dfrac{\sin (6x)}{2} + (\sin(4x)+\sin(2x))+(\sin(5x)+\sin(x))+ \sin(3x)) dx

Simply integrating we get: 81 30 \dfrac{81}{30}

Final result is = 81 30 − 1 6 = 38 15 =\dfrac{81}{30} - \dfrac{1}{6}=\dfrac{38}{15}

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Fixed up your LaTeX: about half the time you used sin or cos you missed the \ in front.

Jason Dyer Staff - 4 years, 6 months ago
Ankush Tiwari
Dec 14, 2016

sin ⁡ ( 3 x ) sin ⁡ ( 5 x 2 ) sin ⁡ ( x 2 ) \dfrac{\sin\left(3x\right)\sin\left(\frac{5x}{2}\right)}{\sin\left(\frac{x}{2}\right)} = 2 sin ⁡ ( 3 x 2 ) cos ⁡ ( 3 x 2 ) sin ⁡ ( 5 x 2 ) sin ⁡ ( x 2 ) = \dfrac{2\sin(\frac{3x}{2})\cos(\frac{3x}{2})\sin(\frac{5x}{2})}{\sin(\frac{x}{2})}

= 2 ( 3 sin ⁡ ( x 2 ) − 4 sin ⁡ ( x 2 ) 3 ) cos ⁡ ( 3 x 2 ) sin ⁡ ( 5 x 2 ) sin ⁡ ( x 2 ) = \dfrac{2(3\sin(\frac{x}{2})-4\sin(\frac{x}{2})^3)\cos(\frac{3x}{2})\sin(\frac{5x}{2})}{\sin(\frac{x}{2})}

= 2 ( 3 − 4 sin ⁡ ( x 2 ) 2 ) cos ⁡ ( 3 x 2 ) sin ⁡ ( 5 x 2 ) = 2(3-4\sin(\frac{x}{2})^2)\cos(\frac{3x}{2})\sin(\frac{5x}{2})

= ( 1 + 2 cos ⁡ x ) ( 2 cos ⁡ ( 3 x 2 ) sin ⁡ ( 5 x 2 ) ) = (1+2\cos x)(2\cos(\frac{3x}{2})\sin(\frac{5x}{2}))

= ( 1 + 2 cos ⁡ x ) ( sin ⁡ 4 x + sin ⁡ x ) = (1+2\cos x)(\sin 4x+\sin x)

= sin ⁡ 4 x + sin ⁡ x + 2 sin ⁡ 4 x cos ⁡ x + 2 sin ⁡ x cos ⁡ x = \sin 4x + \sin x + 2\sin 4x \cos x + 2\sin x\cos x

= sin ⁡ 4 x + sin ⁡ x + sin ⁡ 5 x + sin ⁡ 3 x + sin ⁡ 2 x = \sin 4x + \sin x + \sin 5x + \sin 3x + \sin 2x

⇒ ∫ 0 π 2 sin ⁡ ( 3 x ) sin ⁡ ( 5 x 2 ) sin ⁡ ( x 2 ) d x \Rightarrow \large \int_{0}^{\frac{\pi}{2}}\dfrac{\sin\left(3x\right)\sin\left(\frac{5x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \, dx

= ∫ 0 π 2 ( sin ⁡ 4 x + sin ⁡ x + sin ⁡ 5 x + sin ⁡ 3 x + sin ⁡ 2 x ) d x = 38 15 = \large \int_{0}^{\frac{\pi}{2}}( \sin4x + \sin x + \sin5x + \sin3x + \sin2x)\, dx = \frac{38}{15}

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