A calculus problem by Srinivasa Gopal

Calculus Level 2

0 π 2 sin x sin x + cos x d x = ? \large \int_0^{\tfrac\pi2} \dfrac{\sqrt{\sin x}}{ \sqrt{\sin x} + \sqrt{\cos x}} \, dx = \, ?

π 3 \frac\pi3 π 6 \frac\pi6 π 4 \frac\pi4 π 2 \frac\pi2

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2 solutions

Chew-Seong Cheong
Aug 31, 2020

I = 0 π 2 sin x sin x + cos x d x By reflection a b f ( x ) d x = a b f ( a + b x ) d x = 1 2 0 π 2 ( sin x sin x + cos x + cos x cos x + sin x ) d 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{aligned} I & = \int_0^\frac \pi 2 \frac {\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx & \small \blue{\text{By reflection }\int_a^b f(x) \ dx = \int_a^b f(a+b-x)\ dx} \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac {\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} + \frac {\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac {\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx \\ & = \frac 12 \int_0^\frac \pi 2 dx = \boxed {\frac \pi 4} \end{aligned}


Reference: Reflections

Wow, I didn't even know that. Thanks!

Anh Khoa Nguyễn Ngọc - 9 months, 2 weeks ago

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Yes, it is call "reflections".

Chew-Seong Cheong - 9 months, 2 weeks ago
Srinivasa Gopal
Aug 30, 2020

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