Mother of god, what is that?

Calculus Level 5

E = ∫ 1 / 2 1 2 x sin ⁡ ( 2 tan ⁡ ( x 2 ) ) sin ⁡ 2 ( tan ⁡ ( x 2 ) ) cos ⁡ 2 ( x 2 ) d x \mathfrak{E} = \large \displaystyle \int_{1/2}^{1} \dfrac{2x \sin(2 \tan(x^2 ))}{\sin^2 (\tan(x^2)) \cos^2 (x^2)} \, dx

Find ⌊ 1000 E ⌋ \left \lfloor 1000 \mathfrak{E} \right \rfloor .

Details and assumptions:

After you have integrated and are ready to put in the limits of integration, of course by hand, then you may use a calculator to compute the final answer.


The answer is 2751.

Hobart Pao by Hobart Pao , 23, USA

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1 solution

Rudraksh Shukla
May 7, 2016

Given, ∫ 1 2 1 sin ⁡ ( 2 tan ⁡ ( x 2 ) ) 2 x d x sin ⁡ 2 ( tan ⁡ ( x 2 ) ) cos ⁡ 2 x 2 \int_{\frac{1}{2}}^{1} \dfrac{\sin(2\tan(x^2))2xdx}{\sin^2(\tan(x^2))\cos^2x^2} Using the relations sin ⁡ 2 θ = 2 sin ⁡ θ cos ⁡ θ , cos ⁡ θ = 1 sec ⁡ θ \sin2\theta=2\sin\theta\cos\theta, \cos\theta=\dfrac{1}{\sec\theta} The given integral now is, 2 ∫ 1 2 1 cos ⁡ ( tan ⁡ ( x 2 ) ) sec ⁡ 2 ( x 2 ) 2 x d x sin ⁡ ( tan ⁡ ( x 2 ) ) {2}\int_{\frac{1}{2}}^{1} \dfrac{\cos(\tan(x^2))\sec^2(x^2)2xdx}{\sin(\tan(x^2))} Now substitute as sin ⁡ ( tan ⁡ ( x 2 ) ) = t \sin(\tan(x^2))=t then clearly by chain rule cos ⁡ ( tan ⁡ ( x 2 ) ) sec ⁡ 2 ( x 2 ) 2 x d x = d t \cos(\tan(x^2))\sec^2(x^2)2xdx=dt The above integral then is 2 ∫ d t t 2\int \dfrac{dt}{t} which is 2 l n ( sin ⁡ ( tan ⁡ ( x 2 ) ) ) 2ln(\sin(\tan(x^2))) and substitute the limits.

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