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Edith Mayer

Edith Mayer

Answered question

2022-04-10

Integral d x sin x + sec x

Answer & Explanation

Braxton Gallagher

Braxton Gallagher

Beginner2022-04-11Added 21 answers

I = 1 sin x + sec x d x = cos x sin x cos x + 1 d x = ( cos x + sin x ) + ( cos x sin x ) sin 2 x + 2 d x
= cos x + sin x sin 2 x + 2 d x + cos x sin x sin 2 x + 2 d x = I 1 + I 2
for I 1 = cos x + sin x 3 ( sin x cos x ) 2 d x , put ( sin x cos x ) = t then ( cos x + sin x ) d x = d t
for I 2 = cos x sin x ( sin x + cos x ) 2 + 1 d x , put ( sin x + cos x ) = u then ( cos x sin x ) d x = d u
So I = 1 3 t 2 d t + 1 1 + u 2 d u

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