Integrate cos 3x over sin squared 3x?

Answer 1

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)