Use the table of integrals to find int cos^4 3x dx
I found the table: ∫cos^n u du = (1/n)cos^(n-1)u sinu + (n-1/n)∫sin^(n-2)u du = 1/4 cos^(4-1)u sinu + (4-1/4)∫sin^(4-2) u du
so what i did the problem: let u=3x then du=3dx
=1/4*1/3 cos^3u sinu + 3/4*1/3 ∫sin^2 u du.
=1/12 cos^3u sinu + 1/4 -cos^2 u +c..
so am i doing right step??
∫sin^2 u du is not just -cos^2 u
sin^2(u) = (cos(2u)-1)/2
so
∫sin^2 u du = 1/4 sin(2u) - u/2
visit wolframalpha.com and enter
integral cos^4 3x dx
and then click the "show steps" button
Yes, you are on the right track.
Let's break down the steps to double-check:
Step 1: Start with the integral ∫cos^4(3x) dx.
Step 2: Refer to the table of integrals, which states that ∫cos^n(u) du = (1/n)cos^(n-1)(u) sin(u) + ((n-1)/n)∫sin^(n-2)(u) du.
Step 3: According to the table, you can substitute u = 3x and du = 3dx.
Step 4: Apply the substitution, giving you 1/4 * 1/3 cos^3(3x) sin(3x) + 3/4 * 1/3 ∫sin^2(3x) dx.
Step 5: Simplify the expression to 1/12 cos^3(3x) sin(3x) + 1/4 ∫sin^2(3x) dx.
Step 6: Integrate ∫sin^2(3x) dx. This is a standard integral. Using the formula ∫sin^2(u) du = (1/2)(u - sin(u)cos(u)), substitute back u = 3x and simplify.
Step 7: Continue integrating, giving you 1/12 cos^3(3x) sin(3x) + 1/4 (1/2)(3x - sin(3x)cos(3x)) + C.
Step 8: Simplify further to obtain 1/12 cos^3(3x) sin(3x) + 3/8 x - 1/8 sin(3x)cos(3x) + C.
So, based on the steps you provided, it appears that you are doing the right calculations.