Intermediate integration

Calculus Level 3

Find the value of: ⌊ 1000 ∫ 0 π 2 sin ⁡ x d x sin ⁡ x + cos ⁡ x ⌋ \left\lfloor1000\int_0^{\frac\pi2}\frac{\sin x\mbox{ d}x}{\sin x+\cos x}\right\rfloor

Note : I take no credit.


The answer is 785.

Kenny Lau by Kenny Lau , 22, Hong Kong

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1 solution

Kenny Lau
Jul 11, 2014

I = ∫ 0 π 2 sin ⁡ x d x sin ⁡ x + cos ⁡ x I = ∫ π 2 0 sin ⁡ ( 9 0 ∘ − u ) d ( 9 0 ∘ − u ) sin ⁡ ( 9 0 ∘ − u ) cos ⁡ ( 9 0 ∘ − u ) = ∫ 0 π 2 cos ⁡ u d u sin ⁡ u + cos ⁡ u = ∫ 0 π 2 cos ⁡ x d x sin ⁡ x + cos ⁡ x I = 1 2 ( ∫ 0 π 2 sin ⁡ x d x sin ⁡ x + cos ⁡ x + ∫ 0 π 2 cos ⁡ x d x sin ⁡ x + cos ⁡ x ) = 1 2 ∫ 0 π 2 sin ⁡ x + cos ⁡ x sin ⁡ x + cos ⁡ x d x = 1 2 ∫ 0 π 2 d x = π 4 \begin{array}{rcl} I&=&\int_0^{\frac\pi2}\frac{\sin x\mbox{ d}x}{\sin x+\cos x}\\ I&=&\int_{\frac\pi2}^0\frac{\sin(90^\circ-u)\mbox{ d}(90^\circ-u)}{\sin(90^\circ-u)\cos(90^\circ-u)}\\ &=&\int_0^{\frac\pi2}\frac{\cos u\mbox{ d}u}{\sin u+\cos u}\\ &=&\int_0^{\frac\pi2}\frac{\cos x\mbox{ d}x}{\sin x+\cos x}\\ I&=&\frac12\left(\int_0^{\frac\pi2}\frac{\sin x\mbox{ d}x}{\sin x+\cos x}+\int_0^{\frac\pi2}\frac{\cos x\mbox{ d}x}{\sin x+\cos x}\right)\\ &=&\frac12\int_0^{\frac\pi2}\frac{\sin x+\cos x}{\sin x+\cos x}\mbox{ d}x\\ &=&\frac12\int_0^{\frac\pi2}\mbox{ d}x\\ &=&\frac\pi4 \end{array}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Just a small addition: There is a property of definite integrals: ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x \displaystyle \int \limits_0^a f(x) \text{d}x = \int \limits_0^a f(a-x) \text{d}x
It makes the steps 2 and 3 more elegant. :)

Otherwise, a nice solution!!

Sudeep Salgia - 6 years, 11 months ago

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