Integration of 1/sinx-sin2x dx

Shirley Peck

Shirley Peck

Answered question

2023-03-23

Integration of 1/sinx-sin2x dx

Answer & Explanation

nickalasaurus6ea0

nickalasaurus6ea0

Beginner2023-03-24Added 5 answers

Step 1
1 sin x sin 2 x = d x sin 2 x sin x = sin x d x ( 2 cos x 1 ) ( cos 3 x 1 )
Step 2
Let u=cos x, du=-sin x dx
= d u ( 2 u 1 ) ( u 2 1 ) = d u ( u 1 ) ( u + 1 ) ( 2 u 1 ) = [ u 3 ( 2 u 1 ) + 1 6 ( u + 1 ) + 1 2 ( u 1 ) d u ] = ( 2 3 ln ( 2 u 1 ) + ln ( u + 1 ) 6 + ln ( u 1 ) 2 ) + c = 2 3 ln ( 2 cos x 1 ) ln ( cos x + 1 ) 6 ln ( cos x 1 ) 2 + c

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