Solve it showing detail steps..please.
Integrate : ∫(cos^3 x)^2 dx
∫(cos^3 x)^2 dx
= ∫cos^6 x dx
Now, just as cos(6x) can be expanded into a polynomial in cos(x), cos^6(x) can be expanded into a sum of cosines of multiple angles:
= ∫(1/32)(15cos(2x)+8cos(4x)+cos(6x)+10) dx
= 1/32 (15/2 sin(2x) + 2sin(4x) + 1/6 sin(6x) + 10x) + C
Or, you can use the power reduction formula as shown here
http://www.math-prof.com/Calculus_2/Calc_Ch_07.asp
which are derived using integration by parts:
∫cos^6 x dx
u = cos^5 x, du = -5cos^4 x dx
dv = cosx dx, v = sinx
∫cos^6 x dx = cos^5x sinx + 5∫cos^4x sinx dx
Now it's easy, since you have
u = cosx
du = -sinx dx
oops. du = -5cos^4x sinx dx
things are a bit more complicated than I showed here, since you wind up with a sin^2(x), which has to be converted into (1-cos^2 x) and you go around again.
To integrate ∫(cos^3 x)^2 dx, let's break it down step by step:
Step 1: Expand the function.
Recall the identity for cos^2 x: cos^2 x = (1/2)(1 + cos 2x). Therefore, we can write:
(cos^3 x)^2 = (cos^2 x)^3 = [(1/2)(1 + cos 2x)]^3
Step 2: Simplify the expression.
To simplify, we'll multiply the expression [(1/2)(1 + cos 2x)]^3 by itself.
[(1/2)(1 + cos 2x)]^3 = [(1 + cos 2x)]^3 * (1/2)^3
= (1 + cos 2x)^3 / 8
Step 3: Expand the expression using the binomial theorem.
Applying the binomial theorem, we can express (1 + cos 2x)^3 as:
(1 + cos 2x)^3 = 1^3 + 3(1^2)(cos 2x) + 3(1)(cos 2x)^2 + (cos 2x)^3
= 1 + 3cos 2x + 3(cos^2 2x) + (cos^2 2x)(cos 2x)
Step 4: Simplify further.
Now we have:
∫(cos^3 x)^2 dx = ∫(1 + 3cos 2x + 3(cos^2 2x) + (cos^2 2x)(cos 2x))/8 dx
= (1/8) ∫(1 + 3cos 2x + 3cos^2 2x + cos^3 2x) dx
= (1/8) [∫1 dx + ∫3cos 2x dx + ∫3cos^2 2x dx + ∫cos^3 2x dx]
Step 5: Integrate each term.
∫1 dx = x + C1
∫3cos 2x dx = (3/2) sin 2x + C2
∫3cos^2 2x dx = (3/2) (x/2 + 1/4 sin 4x) + C3
∫cos^3 2x dx is a bit more tricky. We'll need to use a substitution.
Let's assume u = cos 2x. Then du/dx = -2sin 2x, or du = -2sin 2x dx.
Rearranging, dx = -0.5 * du/sin 2x.
Substituting into the integral:
∫cos^3 2x dx = ∫cos^3 2x (-0.5 * du/sin 2x)
= -0.5 ∫cos^3 2x du/sin 2x
= -0.5 ∫(1 - sin^2 2x) du/sin 2x
= -0.5 ∫(1/u^2 - u^2) du
= -0.5 [-1/u - (u^3)/3] + C4
= -0.5 [-(1/cos 2x) - (cos^3 2x)/3] + C4
Step 6: Combine all the terms.
Putting it all together, we have:
∫(cos^3 x)^2 dx = (1/8) [ x + (3/2) sin 2x + (3/2) (x/2 + 1/4 sin 4x) - 0.5 (1/cos 2x + (cos^3 2x)/3)] + C
That's the detailed step-by-step solution to integrating ∫(cos^3 x)^2 dx.