Throwback to 2014

Calculus Level 3

∫ − 2014 2014 d x 1 + sin ⁡ 2015 x + 1 + sin ⁡ 4030 x = ? \large \int ^{2014}_{-2014} \frac {dx}{1+\sin ^{2015}x+\sqrt {1+\sin ^{4030}x}}=?


The answer is 2014.

Parth Sankhe by Parth Sankhe , 27, India

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1 solution

I = ∫ − 2014 2014 d x 1 + sin ⁡ 2015 x + 1 + sin ⁡ 4030 x Using ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x = 1 2 ∫ − 2014 2014 ( 1 1 + sin ⁡ 2015 x + 1 + sin ⁡ 4030 x + 1 1 − sin ⁡ 2015 x + 1 + sin ⁡ 4030 x ) d x = 1 2 ∫ − 2014 2014 1 − sin ⁡ 2015 x + 1 + sin ⁡ 4030 x + 1 + sin ⁡ 2015 x + 1 + sin ⁡ 4030 x ( 1 + sin ⁡ 2015 x + 1 + sin ⁡ 4030 x ) ( 1 − sin ⁡ 2015 x + 1 + sin ⁡ 4030 x ) d x = 1 2 ∫ − 2014 2014 2 + 2 1 + sin ⁡ 4030 x 2 + 2 1 + sin ⁡ 4030 x d x = 1 2 ∫ − 2014 2014 d x = 1 2 x ∣ − 2014 2014 = 2014 \begin{aligned} I & = \int_{-2014}^{2014} \frac {dx}{1+\sin^{2015}x + \sqrt{1+\sin^{4030}x}} & \small \color{#3D99F6} \text{Using }\int_a^b f(x)\ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_{-2014}^{2014} \left(\frac 1{1+\sin^{2015}x + \sqrt{1+\sin^{4030}x}} + \frac 1{1-\sin^{2015}x + \sqrt{1+\sin^{4030}x}} \right) dx \\ & = \frac 12 \int_{-2014}^{2014} \frac {1-\sin^{2015}x + \sqrt{1+\sin^{4030}x} + 1+\sin^{2015}x + \sqrt{1+\sin^{4030}x}} {\left(1+\sin^{2015}x + \sqrt{1+\sin^{4030}x} \right) \left(1-\sin^{2015}x + \sqrt{1+\sin^{4030}x}\right)} dx \\ & = \frac 12 \int_{-2014}^{2014} \frac {2 + 2\sqrt{1+\sin^{4030}x}}{2+2 \sqrt{1+\sin^{4030}x}} dx \\ & = \frac 12 \int_{-2014}^{2014} dx = \frac 12 x \ \bigg|_{-2014}^{2014} = \boxed{2014} \end{aligned}

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