Integration problem

Calculus Level 3

I = 0 π 4 sin ( x ) d x + 0 π 4 sin 2 ( x ) d x + 0 π 4 sin 3 ( x ) d x + 0 π 4 sin 4 ( x ) d x + I=\int_0^\frac \pi 4 \sin \left( x\right) dx +\int_0^\frac \pi 4 \sin ^{2}\left( x\right) dx + \int_0^\frac \pi 4 \sin ^{3}\left( x\right) dx + \int_0^\frac \pi 4\sin ^{4}\left( x\right) dx+\cdots Given that I = A π B I=\sqrt{A} - \dfrac{\pi}{B} , find A + B A+B .


The answer is 6.

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1 solution

Rishabh Jain
Jan 29, 2016

I = 0 π 4 sin x + sin 2 x + sin 3 x + . . . \large I=\int_{0}^\frac{\pi}{4} \sin x +\sin^2 x+ \sin^3 x+... Since |sin x| \leq 1, integral is the sum of an infinite G.P whose first term as well as common difference is sin x \color{#EC7300}{\sin x}
I = 0 π 4 sin x 1 sin x \therefore I=\int_{0}^{\frac{\pi}{4}}\dfrac{\sin x}{1-\sin x} = 0 π 4 sec x tan x + tan 2 x =\int_{0}^{\frac{\pi}{4}} \sec x \tan x +\tan^2x = sec x + tan x x 0 π 4 =\sec x +\tan x -x |_{\small 0}^{\small{ \frac{\pi}{4}}} = 2 π 4 =\large \sqrt2-\dfrac{\pi}{4} 4 + 2 = 6 \Large 4+2=\boxed{\color{#20A900}{\boxed{6}}}

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