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integration ofsin^6x cos^4x dx

 integration ofsin^6x cos^4x dx

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1 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
8 years ago
sin2A = 2sinA cosA
cos2A = 2cos²A - 1 = 1 - 2sin²A
∫ sinx dx = -cosx + C
∫ cosx dx = sinx + C

∫ sin⁴x cos⁶x dx
= ∫ sin⁴x cos⁴x cos²x dx
= (1/16) ∫ 16 sin⁴x cos⁴x cos²x dx
= (1/16) ∫ (2 sinx cosx)⁴ [ (cos2x + 1)/2] dx
= (1/32) ∫ (2 sinx cosx)⁴ (cos2x + 1) dx
= (1/32) ∫ (sin2x)⁴ (cos2x + 1) dx
= (1/32) ∫ (sin⁴2x cos2x + sin⁴2x) dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/32) ∫ sin⁴2x dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/32) ∫ (sin²2x)² dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/32) ∫ [ (1 - cos4x)/2 ]² dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/32) ∫ (1 - cos4x)²/4 dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/128) ∫ (1 - cos4x)² dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/128) ∫ (cos²4x - 2cos4x + 1) dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/128) ∫ cos²4x dx - (1/64) ∫ cos4x dx + (1/128) ∫ dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/128) ∫ [(cos8x + 1)/2] dx - (1/64) ∫ cos4x dx + (1/128) ∫ dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/256) ∫ (cos8x + 1) dx - (1/64) ∫ cos4x dx + (1/128) ∫ dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/256) ∫ cos8x dx + (1/256) ∫ dx - (1/64) ∫ cos4x dx + (1/128) ∫ dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/256) ∫ cos8x dx - (1/64) ∫ cos4x dx + (3/256) ∫ dx

the only problem here is to solve the first integral in the above expression
∫ sin⁴2x cos2x dx
let u = sin2x ==> du/dx = 2cos2x ==> dx = du/(2cos2x)

∫ sin⁴2x cos2x dx
= ∫ u⁴ cos2x [du/(2cos2x)]
= (1/2) ∫ u⁴ du
= (1/2)(1/5) u⁵ + C
= (1/10) u⁵ + C
= (1/10) sin⁵2x + C

The finishing step:
∫ sin⁴x cos⁶x dx
= (1/32) ∫ sin⁴2x cos2x dx + (1/256) ∫ cos8x dx - (1/64) ∫ cos4x dx + (3/256) ∫ dx
= (1/32) (1/10) (sin⁵2x) + (1/256) (1/8)(sin8x) - (1/64) (1/4)(sin4x) + (3/256) x + C
= (1/320) sin⁵2x + (1/2048) sin8x - (1/256) sin4x + (3/256) x + C ⇦ ANSWER

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