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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,

âˆ«sinn(u) du = (-1/n)sinn-1(u)cos(u) + ((n-1)/n)âˆ«sinn-2(u) du

where n â‰¥ 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

âˆ«sin7 3x dx = (1/3)âˆ«sin7(u) du

âˆ«sin7(u) du = (-1/7)sin6(u)cos(u) + (6/7)âˆ«sin5(u) du

= (-1/7)sin6(u)cos(u) + (6/7)[(-1/5)sin4(u)cos(u) + (4/5)âˆ«sin3(u) du]

= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) + (24/35)âˆ«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)âˆ«sin(u) du]

= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) + (16/35)âˆ«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

âˆ«sin7 3x dx = (1/3)âˆ«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

âˆ«sin7 3x dx = âˆ«sin63x sin 3x dx = âˆ«(sin2 3x)3 sin 3x dx =âˆ«(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)âˆ«(1 - u2)3 du = âˆ«(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

Q: What is the integral of sine to the 7th power 3x dx?

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X^2? 1/3x^3

The 3s would cancel and it would become the integral of 1/x which is ln x.

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

multiply the number times x. For example, the integral of 3 is 3x.

7

Related questions

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.

-3x

-cos(3x) + constant

The GCF is x5

∫(-3)dx = -3x + C

dy/dx = 3 integral = (3x^2)/2

7

X^2? 1/3x^3

The 3s would cancel and it would become the integral of 1/x which is ln x.

(3x-1) * exp(3*x) / 9

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If you want sin(3x) + cos(3x) = 6, then this is impossible. Sine and cosine will only return values between -1 and 1, so the expression sin(3x) + cos(3x) could only take values from -2 to 2, although even this is to great as sine and cosine of the same number will never both be 1 or -1. Similarly, if you want a solution to sin3x + cos3x = 6, then this is also impossible, because any power of a number between -1 and 1 will itself be between -1 and 1.