three 3's

Calculus Level 4

∣ ⌊ ∫ 0 3 x 3 cos ⁡ ( 3 ln ⁡ x ) dx ⌋ ∣ \displaystyle \Bigg|\bigg\lfloor\int_{0}^{3}x^3 \cos\big(3 \ln x) \text{dx}\bigg\rfloor \Bigg|


The answer is 15.

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Jatin Yadav by jatin yadav , 23, India

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2 solutions

Pi Han Goh
Dec 26, 2013

Let x = e y ⇒ d x = e y d y x = e^y \Rightarrow \mathrm{d} x =e^y \mathrm{d} y

The integral becomes

= ∫ − ∞ ln ⁡ 3 e 4 y cos ⁡ ( 3 y ) d y \int_{- \infty}^{\ln 3} e^{4y} \cos (3y) \mathrm{d} y

Let z = − y z = -y

= ∫ − ln ⁡ 3 ∞ e − 4 z cos ⁡ ( 3 z ) d z \int_{-\ln 3}^{\infty} e^{-4z} \cos (3z) \mathrm{d} z

Solve it like this

With a = − 4 a = -4 , b = 3 b = 3

= − 1 25 ( e − 4 ln ⁡ 3 ) ( − 4 cos ⁡ ( − 3 ln ⁡ 3 ) + 3 sin ⁡ ( − 3 ln ⁡ 3 ) ) ≈ − 14.2995 =- \frac {1}{25} \left ( e^{-4 \ln 3} \right ) \left ( -4 \cos (-3 \ln 3) + 3 \sin (-3 \ln 3) \right ) \approx - 14.2995

Answer is 15 \boxed{15}

14 Helpful 0 Interesting 0 Brilliant 0 Confused

Very nice, mine is almost similar. I took A as the given integral and B as the same integral as A with cosine replaced by sine. I got A + i B = ∫ 0 3 x 3 + 3 i d x A+iB=\int_0^3 x^{3+3i}dx and solving this is easy.

Pranav Arora - 7 years, 5 months ago

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Really nice solution! I took t = ln(x) and integration by parts, I'm just reading that note. Greetings!

Carlos David Nexans - 6 years, 10 months ago

Nicely done π \pi , You could have saved a step by substituting x = e − y x = e^{-y} , i did in a similar way as Pranav.

jatin yadav - 7 years, 5 months ago

I use this: ∫ − ∞ ln ⁡ 3 e 4 y cos ⁡ ( 3 y ) d y = ∫ − ∞ ln ⁡ 3 e 4 y R e ( e 3 i y ) d y = R e ( ∫ − ∞ ln ⁡ 3 e 4 + 3 i y d y ) . \begin{aligned} \int_{-\infty}^{\ln 3} e^{4y} \cos(3y)\,dy &=\int_{-\infty}^{\ln 3} e^{4y}\; \mathbf{Re}\,\;\left( e^{3iy}\right) \,dy\\ &=\mathbf{Re}\,\;\left( \int_{-\infty}^{\ln 3} e^{4+3iy}\,dy\right). \end{aligned}

Tunk-Fey Ariawan - 7 years, 3 months ago

Did the exact same! I don't understand why so much calculations were made to require at the end.

Kartik Sharma - 6 years, 3 months ago
Tunk-Fey Ariawan
Feb 15, 2014

That sign!? It took 3x for me to answer this problem.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Let x = e z \,x=e^z , then d x = e z d z \,dx=e^z\,dz . For 0 ≤ x ≤ 3 \,0\le x \le 3 corresponding to − ∞ ≤ x ≤ ln ⁡ 3 \,-\infty \le x \le \ln 3 . Thus, ∫ 0 3 x 3 cos ⁡ ( ln ⁡ x ) d x = ∫ − ∞ ln ⁡ 3 e 4 z cos ⁡ ( 3 z ) d z = ∫ − ∞ ln ⁡ 3 e 4 z R e ( e 3 i z ) d z = R e ( ∫ − ∞ ln ⁡ 3 e 4 + 3 i z d z ) . \begin{aligned} \int_0^3 x^3 \cos(\ln x)\,dx&= \int_{-\infty}^{\ln 3} e^{4z} \cos(3z)\,dz\\ &=\int_{-\infty}^{\ln 3} e^{4z}\; \mathbf{Re}\,\;\left( e^{3iz}\right) \,dz\\ &=\mathbf{Re}\,\;\left( \int_{-\infty}^{\ln 3} e^{4+3iz}\,dz\right). \end{aligned} The rest would be easy. # Q . E . D . # \text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}

Tunk-Fey Ariawan - 7 years, 3 months ago

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