Question

integral of sin2x/1+sinx​

Answer 1

Answer:

To integrate sin2x, also written as ∫sin2x dx, and sin 2x, we usually use a u substitution to build a new integration in terms of u.

Step-by-step explanation:

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Answer 2

Answer:

Let I =

∫sin 2x(1+sin x)(2+sin x)dx

I =

∫2sin x cos x(1+sin x)(2+sin x)dx

Put sin x = t ⇒ cos x dx = dt

∴ I = ∫2t(1+t)(2+t)dt

Let

2t(1+t)(2+t)=At+1+Bt+2

⇒ 2t = A(t + 2) + B(t + 1)

Put t + 2 = 0 ⇒ t = -2

- 4 = B (-2 + 1) ⇒ B = 4

Put t + 1 = 0 ⇒ t = -1

-2 = A (-1 + 2) ⇒ A = -2

∴I=∫[-2t+1+4t+2]dt

= -2 log (t + 1) + 4 log (t + 2) + C

= -2 log (sin x + 1) + 4 log (sin x + 2) + C