Hello,
∫ sin⁵x √(cosx) dx =
let's rewrite the integral as:
∫ [(sin²x)² √(cosx)] sinx dx =
let's apply the identity sin²x = 1 - cos²x:
∫ [(1 - cos²x)² √(cosx)] sinx dx =
(expanding the square)
∫ [(1 - 2cos²x + cos⁴x) √(cosx)] sinx dx =
let:
√(cosx) = u
cosx = u²
(differentiating both sides)
d(cosx) = d(u²) → - sinx dx = 2u du → sinx dx = - 2u du
then, substituting:
∫ [(1 - 2cos²x + cos⁴x) √(cosx)] sinx dx = ∫ [1 - 2(u²)² + (u²)⁴] u (- 2u) du =
∫ (1 - 2u⁴ + u⁸) (- 2u²) du =
∫ (- 2u² + 4u⁶ - 2u¹º) du =
(splitting into three integrals and pulling constants out)
- 2 ∫ u² du + 4 ∫ u⁶ du - 2 ∫ u¹º du =
- 2 [1/(2+1)] u^(2+1) + 4 [1/(6+1)] u^(6+1) - 2 [1/(10+1)] u^(10+1) + C =
- (2/3)u³ + (4/7)u⁷ - (2/11)u¹¹ + C
let's substitute back √(cosx) for u, ending with:
∫ sin⁵x √(cosx) dx = - (2/3)√(cosx)³ + (4/7)√(cosx)⁷ - (2/11)√(cosx)¹¹ + C
I hope it's helpful
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Verified answer
Hello,
∫ sin⁵x √(cosx) dx =
let's rewrite the integral as:
∫ [(sin²x)² √(cosx)] sinx dx =
let's apply the identity sin²x = 1 - cos²x:
∫ [(1 - cos²x)² √(cosx)] sinx dx =
(expanding the square)
∫ [(1 - 2cos²x + cos⁴x) √(cosx)] sinx dx =
let:
√(cosx) = u
cosx = u²
(differentiating both sides)
d(cosx) = d(u²) → - sinx dx = 2u du → sinx dx = - 2u du
then, substituting:
∫ [(1 - 2cos²x + cos⁴x) √(cosx)] sinx dx = ∫ [1 - 2(u²)² + (u²)⁴] u (- 2u) du =
∫ (1 - 2u⁴ + u⁸) (- 2u²) du =
∫ (- 2u² + 4u⁶ - 2u¹º) du =
(splitting into three integrals and pulling constants out)
- 2 ∫ u² du + 4 ∫ u⁶ du - 2 ∫ u¹º du =
- 2 [1/(2+1)] u^(2+1) + 4 [1/(6+1)] u^(6+1) - 2 [1/(10+1)] u^(10+1) + C =
- (2/3)u³ + (4/7)u⁷ - (2/11)u¹¹ + C
let's substitute back √(cosx) for u, ending with:
∫ sin⁵x √(cosx) dx = - (2/3)√(cosx)³ + (4/7)√(cosx)⁷ - (2/11)√(cosx)¹¹ + C
I hope it's helpful