0
If you look in most calculus books there is a table of intergals. Your problem is one of the trigonometric forms. Below is given the reduction formula,
∫sinn(u) du = (-1/n)sinn-1(u)cos(u) + ((n-1)/n)∫sinn-2(u) du
where n ≥ 2 is an integer.
Basically yours would be n=7 & u=3x. You keep integrating until you lose the last integral sign on the right side. Pretty tedious work, but if I am using the right integral form & didn't make any mistakes the answer should be:
(-1/7)sin6(3x)cos(3x) - (6/35)sin4(3x)cos(3x) - (8/35)sin2(3x)cos(3x) - (16/35)cos(3x) + C
Here is the work:
Let u = 3x
u' = 3 dx, so that dx = du/3
∫sin7 3x dx = (1/3)∫sin7(u) du
∫sin7(u) du = (-1/7)sin6(u)cos(u) + (6/7)∫sin5(u) du
= (-1/7)sin6(u)cos(u) + (6/7)[(-1/5)sin4(u)cos(u) + (4/5)∫sin3(u) du]
= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) + (24/35)∫sin3(u) du
= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) + (24/35)[(-1/3)sin2(u)cos(u) + (2/3)∫sin(u) du]
= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) + (16/35)∫sin(u) du
= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) - (16/35)cos(u) + C
so that
∫sin7 3x dx = (1/3)∫sin7(u) du
= (1/3 [(-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) - (16/35)cos(u)] + C
= -(1/21)sin6(u)cos(u) - (2/35)sin4(u)cos(u) - (8/105)sin2(u)cos(u) - (16/105)cos(u) + C
= -(1/21)sin6(3x)cos(3x) - (2/35)sin4(3x)cos(3x) - (8/105)sin2(3x)cos(3x) - (16/105)cos(3x) + C
Or
∫sin7 3x dx = ∫sin63x sin 3x dx = ∫(sin2 3x)3 sin 3x dx =∫(1 - cos2 3x)3 sin 3x dx
Let u = cos 3x
u' = (cos 3x)'
du = -sin 3x*3 dx, and dx = du/-3sin 3x
= (-1/3)∫(1 - u2)3 du = ∫(1 - 3u2 + 3u4 + u6) du = (-1/3) [u - (3/3)u3 + (3/5)u5 - u7/7] + C
= (-1/3)u + (1/3)u3 - (1/5)u5 + (1/21)u7 + C
= (-1/3)cos 3x + (1/3)cos3 3x - (1/5)cos5 3x + (1/21)cos7 3x + C
Wiki User
∙ 13y ago
Add your answer:
Earn +20 pts
What is the integral of expx2?
X^2? 1/3x^3
What is the integral of 3 over 3x?
The 3s would cancel and it would become the integral of 1/x which is ln x.
Parentheses before -1 and after 3x and before 3 and after 3x What is the answer to -2 cos2 3x-3 sin 3x equals 0?
I think you mean to solve:(-2 cos23x) - (3 sin 3x) = 0cos2 x + sin2 x = 1⇒ cos2x = 1 - sin2 x⇒ -2 cos2 3x - 3 sin 3x = -2(1 - sin2 3x) - 3 sin 3x = 0⇒ 2 sin2 3x - 3 sin 3x - 2 = 0⇒ (2 sin 3x + 1)(sin 3x - 2) = 0⇒ sin 3x = -1/2 or 2sin 3x = 2 is impossible as the range of sine is -1 ≤ sine ≤ 1Thus:sin 3x = -1/2⇒ 3x = 2nπ - π/6 or 2nπ - 5π/6⇒ x = 2/3nπ -π/18 or 2/3nπ -5π/18
How do you integrate a number in respect to x?
multiply the number times x. For example, the integral of 3 is 3x.
What is 3x to the -4 power?
7
Integral of 3 raised to x?
The integral of 3x is ln(3)*3x. Take the natural log of the base and multiply it by the base raised to the power.
Integral of -3 dx?
-3x
What is the integral of 3sin3x?
-cos(3x) + constant
What is the greatest common factor 19x to the 7th power and 3x to the 5th power?
The GCF is x5
What is the integral of -3dx?
∫(-3)dx = -3x + C
How do you Differentiate and Integrate y equals 3x?
dy/dx = 3 integral = (3x^2)/2
Can 3x-4 be simplified?
7
What is the integral of expx2?
X^2? 1/3x^3
What is the integral of 3 over 3x?
The 3s would cancel and it would become the integral of 1/x which is ln x.
What is integral of xe3x?
(3x-1) * exp(3*x) / 9
Integral of 3x-2 dx?
5
What is the degree of the following polynomial 3x to the seventh power plus 4 x to the 6th power minus 2x to the 4th power plus 8?
A 7th degree polynomial.
