Integrate cos⁴ x ÷ sin^7
Integrate cos⁴ x/sin^7 x dx
BOTH of your solutions are wrong!
cos⁴ x ÷ sin^7 x ≠ (cos² x)² ÷ (sin² x)³ (cos² x)
and the u=sinx substitution is also wrong, since
u = sinx and so du = cosx dx
∫ cos⁴ x/sin^7 x dx = ∫ cos^3 x / sin^7 x * cosx dx
= ∫ (1-u^2)^(3/2) / u^7 du
I apologize for the mistakes in my previous solutions. Thank you for correcting them.
To integrate cos⁴ x ÷ sin^7 x, we can rewrite the expression as:
cos⁴ x ÷ sin^7 x = cos³ x ÷ sin^6 x * cos x ÷ sin x
Then, using the substitution u = sin x, we get:
∫ cos⁴ x ÷ sin^7 x dx = ∫ (1 - u²)³ ÷ u^6 * cos x ÷ sin x dx
Using integration by parts with u = cos x and dv = (1 - u²)³ ÷ u^6 dx, we get:
∫ (1 - u²)³ ÷ u^6 * cos x ÷ sin x dx = [(1 - u²)³ ÷ u^5] * sin x + 3∫ [(1 - u²)² ÷ u^4] sin x dx
Using integration by parts again with u = (1 - u²)² and dv = sin x dx, we get:
∫ (1 - u²)³ ÷ u^6 * cos x ÷ sin x dx = [(1 - u²)³ ÷ u^5] * sin x + 3[(-1/3)(1 - u²)³ ÷ u^3 cos x - 2/3∫ (1 - u²)² ÷ u^2 cos x dx]
Substituting u = sin x back into the equation and simplifying the integral, we get:
∫ cos⁴ x ÷ sin^7 x dx = -cos x ÷ (3sin³ x) - 2cos x ÷ (3sin³ x) - 4/3∫ (1 - u²) ÷ u^2 dx
= -cos x ÷ (sin³ x) - 4/3 * ln |sin x| + C
Therefore, the final solution to the integral of cos⁴ x ÷ sin^7 x is:
∫ cos⁴ x ÷ sin^7 x dx = -cos x ÷ (sin³ x) - 4/3 * ln |sin x| + C
To integrate cos⁴(x) ÷ sin^7(x), we can start by simplifying the expression using trigonometric identities.
Using the Pythagorean identity, sin²(x) + cos²(x) = 1, we can express cos⁴(x) in terms of sin(x):
cos⁴(x) = (1 - sin²(x))² = (1 - sin²(x)) * (1 - sin²(x)) = (1 - sin²(x)) * (1 + sin(x)) * (1 - sin(x))
Now, let's rewrite the expression:
cos⁴(x) ÷ sin^7(x) = [(1 - sin²(x)) * (1 + sin(x)) * (1 - sin(x))] ÷ sin^7(x)
Using the properties of exponents, we can rewrite the expression as:
cos⁴(x) ÷ sin^7(x) = (1 - sin²(x)) ÷ sin^5(x) * (1 + sin(x)) * (1 - sin(x))
Now, we can split the expression into two separate fractions:
cos⁴(x) ÷ sin^7(x) = [1 ÷ sin^5(x)] * [(1 - sin²(x)) * (1 + sin(x)) * (1 - sin(x))]
Next, let's simplify each fraction:
1 ÷ sin^5(x) = csc^5(x)
(1 - sin²(x)) = cos²(x)
(1 + sin(x)) * (1 - sin(x)) = 1 - sin²(x) = cos²(x)
Now, our expression becomes:
cos⁴(x) ÷ sin^7(x) = [csc^5(x)] * [(cos²(x)) * (cos²(x))]
Finally, using the power rule of integration, we can integrate each part separately:
∫ [csc^5(x)] * [(cos²(x)) * (cos²(x))] dx = ∫ csc^5(x) dx * ∫ cos⁴(x) dx
The integral of csc^5(x) can be solved using integration by parts or by using trigonometric identities. The integral of cos⁴(x) can be solved using the power rule of integration.
Considering the complexity of these integrals, it might be a good idea to use a computer algebra system or numerical methods to approximate the integral.
To integrate cos^4(x) / sin^7(x), we can use a combination of trigonometric identities and the technique of substitution.
Let's start by rewriting the expression using the identity cos^2(x) = 1 - sin^2(x):
cos^4(x) / sin^7(x) = (cos^2(x))^2 / sin^7(x)
= (1 - sin^2(x))^2 / sin^7(x)
Next, we can make a substitution to simplify the expression further. Let u = sin(x), so du = cos(x) dx. This substitution allows us to transform the integral from a trigonometric function to a rational function.
When substituting, we have:
(1 - sin^2(x))^2 / sin^7(x) = (1 - u^2)^2 / u^7
Now, the integral becomes:
∫ (1 - u^2)^2 / u^7 du
Expanding the numerator gives:
∫ (1 - 2u^2 + u^4) / u^7 du
We can now split this fraction into three separate integrals:
∫ (1/u^7 - 2u^2/u^7 + u^4/u^7) du
Simplifying further gives:
∫ u^(-7) du - 2 ∫ u^(-5) du + ∫ u^(-3) du
Integrating each term:
(1/-6)u^(-6) - 2(1/-4)u^(-4) + (1/-2)u^(-2) + C
Replacing u back with sin(x) and simplifying gives us the final result:
-1/6sin^(-6)(x) + 1/2sin^(-2)(x) - 1/2sin^(-4)(x) + C
Therefore, the integral of cos^4(x) / sin^7(x) is -1/6sin^(-6)(x) + 1/2sin^(-2)(x) - 1/2sin^(-4)(x) + C, where C is the constant of integration.
First, we can rewrite cos⁴ x as (cos² x)². Then, we can use the identity sin² x + cos² x = 1 to rewrite the expression as:
(cos² x)² ÷ (sin² x)³ (cos² x)
Next, we can replace (cos² x) with 1 - (sin² x) to get:
(1 - sin² x)² ÷ (sin² x)³ (1 - sin² x)
Expanding the numerator, we get:
(1 - 2sin² x + sin⁴ x) ÷ (sin² x)³ (1 - sin² x)
Simplifying the denominator, we get:
sin⁴ x ÷ (sin⁴ x cos² x)
Canceling out a factor of sin⁴ x and replacing cos² x with 1 - sin² x, we get:
1 ÷ (sin⁴ x - sin⁶ x)
Finally, we can rewrite sin⁴ x - sin⁶ x as sin⁴ x (1 - sin² x) and use the identity sin² x + cos² x = 1 to get:
1 ÷ (sin⁴ x cos² x)
Therefore, the final result of integrating cos⁴ x ÷ sin^7 is:
∫ cos⁴ x ÷ sin^7 dx = ∫ 1 ÷ (sin⁴ x cos² x) dx = - 1 ÷ (3 sin³ x cos x) + C, where C is the constant of integration.
We can start by simplifying the expression using trigonometric identities. First, we can express cos⁴ x as (cos² x)², and then use the identity sin² x + cos² x = 1 to rewrite the denominator as (sin² x)⁴ sin x. Thus, the integral becomes:
∫ cos⁴ x/sin^7 x dx = ∫ cos² x/sin^5 x * cos² x/sin^2 x dx
Using the substitution u = sin x, we can rewrite the integral as:
∫ (1 - u²) / u^5 * (1 - u²) / u^2 du
Expanding the numerator, we get:
∫ (1 - 2u² + u⁴) / u^7 du
Separating the terms and integrating each one, we obtain:
∫ (1/u^7) du - 2∫(1/u^5) du + ∫(1/u^3) du
= -1/(6u^6) + 1/(2u^4) - 1/(2u^2) + C
Substituting back u = sin x, we get:
∫ cos⁴ x/sin^7 x dx = -1/(6sin⁶ x) + 1/(2sin⁴ x) - 1/(2sin² x) + C
Therefore, the integral of cos⁴ x/sin^7 x is:
∫ cos⁴ x/sin^7 x dx = -1/(6sin⁶ x) + 1/(2sin⁴ x) - 1/(2sin² x) + C