Trigonometric integrals. Evaluate the following integrals. \int \sin^{2}x\cos^{2}xdx

Gregory Jones

Gregory Jones

Answered question

2021-12-12

Trigonometric integrals. Evaluate the following integrals.
sin2xcos2xdx

Answer & Explanation

alexandrebaud43

alexandrebaud43

Beginner2021-12-13Added 36 answers

Step 1
We have to evaluate the integral:
sin2xcos2xdx
Since we know the identity, 2sinxcosx=sin2x
Therefore multiplying and dividing by 22,
2222sin2xcos2xdx=1422sin2xcos2xdx
=14(2sinxcosx)2dx
=14(sin2x)2dx
=14sin22xdx
Step 2
We have identity,
cos2xsin2x=cos2x
cos2x=12sin2x(cos2x+sin2x=1)
sin2x=1cos2x2
Therefore replacing x to 2x, we get
sin22x=1cos4x2
Step 3
Substituting above result in the integral,
14sin22xdx=141cos4x2dx
=18(1cos4x)dx
=18(dxcos4xdx)
=18x18(sin4x4)+C
=18(xsin4x4)+C
Where, C is an arbitrary constant.
Hence, value of integral is 18(xsin
jean2098

jean2098

Beginner2021-12-14Added 38 answers

sin(x)2cos(x)2dx
14×sin(2x)2dx
14×sin(2x)2dx
14×sin(t)22dt
14×12×sin(t)2dt
18×1cos(2t)2dt
18×12×1cos(2t)dt
116×(1dtcos(2t)dt)
116×(tsin(2t)2)
116×(2xsin(2×2x)2)
18xsin(4x)32
Add C
Answer:
18xsin(4x)32+C

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