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.
Integrate cos⁴ x ÷ sin^7
5 answers
Integrate cos⁴ x/sin^7 x dx
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
∫ 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
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
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 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