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Simple Integration 1.03

Evaluate the following integral (could get lengthy)

\(\int \frac{\sqrt{cos2x}}{sinx}dx\)

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Note by Anirudha Nayak
3 years, 7 months ago

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I = ∫ [ √( cos 2x ) / ( sin x ) ] dx

= ∫ (1/ sin x ) √[ (1-tan² x) / (1+tan² x) ] dx

= ∫ [ √(1-tan² x) / ( tan x ) ] dx ......................... (1) .........................................

Put ; t² = 1- tan² x.

Then : 2t. dt/dx = -2 tan x. sec² x

dx = -t dt / ( tan x. sec² x ) .........................................

From (1), then,

I = ∫ ( t / tan x ) ( -t dt / tan x sec² x)

= - ∫ { t² / [ tan² x. sec² x ] } dt

= - ∫ { t² / [ ( 1 - t² ) ( 1 + tan² x ) ] } dt

= - ∫ { t² / [ ( 1 - t² )( 1 + (1-t²)) ] } dt

= - ∫ { t² / [ (1 - t²)( 2 - t²) ] } dt

= - ∫ { [ 1 / (1-t²) + [ -2 / (2-t²) ] } dt ... by Partial Fractions

= - (1/2) ln | (1+t) / (1-t) | + 2(1/ 2√2) ln | (√2 + t) / (√2 - t) | + C, ... t = √(1-tan² x)

Mahmoud Hassona - 3 years, 5 months ago

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