integration of sin^2cos^2?

Question

hi everyone im having a really difficult time with this problem so ur help would be greatly appreciated! :)
How would you integrate sin^2cos^2?

thanks in advance

Answer 1

K = ∫ sin² x cos² x dx



K = ∫ (1 - cos² x) cos² x dx



K = ∫ cos² x dx - ∫ cos⁴ x dx



K = ∫ (½ + ½ cos 2x) dx - ∫ (½ + ½ cos 2x)² dx



K = ½ x + ¼ sin 2x - ∫ (¼ + ½ cos 2x + ¼ cos² 2x) dx



K = ½ x + ¼ sin 2x - ∫ (¼ + ½ cos 2x + ¼ (½ + ½ cos 4x)) dx



K = ½ x + ¼ sin 2x - ∫ (¼ + ½ cos 2x + ⅛ + ⅛ cos 4x) dx



K = ½ x + ¼ sin 2x - ¼ x - ¼ sin 2x - ⅛ x - (1/32) sin 4x + C



K = ⅛ x - (1/32) sin 4x + C

Answer 2

I dont think it is 1. this is what u can do

sin^2cos^2 = (sin(2x)^2)/4

so integration of (1/4)(sin(2x)^2)

then use the u substitution 2x

u = 2x

du = 2 dx

substitute u get

(1/8)(sin(u)^2) du then evaluate integration of (sin u)^2 then times (1/8) later. u should be able to evaualte sin(u)^2 easily i believe. after that, dont forget to substitute back the value from u to x to get the correct answer. (u = 2x)

Answer 3

(sin x)^2 + (cos x)^2 = ; (cos x)^2 = 1 - (sin x)^2; (sin x)^2 ( 1 - (sin x)^2) = (sin x)^2 - (sin x)^4 ; so integration of [(sin x)^2 *(cos x)^2 dx] = integration of [(sin x)^2 dx] - integration of [(sin x)^4 dx] ; int[(sin x)^2 dx] = (1/2)x -(1/4)sin 2x + C ; int[ (sin x)^n dx] = -(1/n)(sin x)^(n-1)(cos x) + ( (n-1)/n) integration of (sin x)^(n-2) dx ; int[ (sin x)^4 dx] = -(1/4)(sin x)^3(cos x) + (3/8)x - (3/16)sin 2x ; so (1/4)(sin x)^3(cos x) + (1/8)x -(1/16)sin 2x

Answer 4

integration of {sin^2(x)cos^2(x)}

integration of =1/4 { 4sin^2(x)cos^2(x)}

integration of =1/4 { 2sin(x)cos(x)}^2

= 1/4{sin(2x)^2}

= 1/4{sin^2(2x)}

= 1/8{1 - cos(4x)}

= 1/8 - 1/8{cos(4x)

Now integrate

= (1/8)(x) -(1/8){sin(4x)}/4 +C

= x/8 -(1/32){sin(4x)} +C.......................Ans

Answer 5

sin^2(x)cos^2(x)



= sin^2(x)[1-sin^2(x)]



= sin^2(x) - sin^4(x)



= 1/2 - (1/2)cos(2x) - 1/4 + (1/2)cos(2x) - (1/4)cos^2(2x)



= ∫[1/4 - (1/4)cos^2(2x)]dx



u = 2x



du = 2dx, dx = (1/2)du



= (1/2)∫[1/4 - (1/4)cos^2(u)]du



= (1/2)∫{1/4 - (1/4)[1/2 + (1/2)cos(2u)]}du



= ∫[1/8 - 1/16 - (1/16)cos(2u)]du



= (1/16)u - (1/32)sin(2u) + C



But u = 2x, so



= x/8 - (1/32)sin(4x) + C

Answer 6

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Answer 7

... ∫ ( sin² x cos² x ) dx



= (1/4) ∫ ( 2 sin x cos x )² dx



= (1/4) ∫ sin² 2x dx



= (1/4) ∫ (1/2)[ 1 - cos 2(2x) ] dx



= (1/8) ∫ ( 1 - cos 4x ) dx



= (1/8) [ x - ( sin 4x / 4 ) ] + C



= ( x/8 ) - ( sin 4x / 32 ) + C ................ Ans.

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A Bonus Integral

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... ∫ [ 1 / ( sin² x cos² x ) ] dx



= ∫ [ ( sin² x + cos² x ) / ( sin² x cos² x ) ] dx ... Now Split



= ∫ [ ( 1/ cos² x ) ] + [ 1/ sin² x ) ] dx



= ∫ sec² x dx + ∫ csc² x dx



= tan x - cot x + C ...................... Ans.

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Happy To Help !

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