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okay I'll use a / to represent the fraction and F(x) as the integration sign
F(x)cos(3x)/sin2(3x)dx
- This problem requires substitution
- U=problem ; DU=anti-derivative
U=Sin(3x) ; DU=-(3)Cos(3x)
- note that U is not squared in the substitution
- also note that DU is multiplied by -3
- you will have to balance the above by canceling
- Rewrite the Equation with the substitutions/cancellations in place
- I have put in bold the changes necessary for substitution
F(x)-(1/3)DU/U2dx
- put the -1/3 of the DU behind the integration sign
- separate the fraction in the equation to make it easier
-1/3F(x)1/U2*DUdx
- to eliminate the fraction make the exponent on U negative
-1/3F(x)U-2DUdx
- integrate the function using the formula Nx=[1/(x+1)]N(x+1)
(-1/3) [1/(-2+1)] U(-2+1)
- simplify the equation
(-1/3)[-1]U(-1)
(1/3)U-1
- then plug in U (U=sin(3x))
(1/3)sin-1(3x)
Your Answer Is:
F(x)cos(3x)/sin2(3x)=(1/3)sin-1(3x)
Also written as:
1/[3sin-1(3x)] or -1/(3sin3x)
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∙ 14y ago
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