How do I evaluate this integral: x^9cos(x^5)dx
Hint: First make a substitution and then use integration by parts.
If w = x^5, 5x^4 dx = dw
The integral you want becomes the integral of
(1/5) w cos w dw
Now use integration by parts, with
u = w dv = cosw dw
du = dw v = sin w
The integral becomes
(1/5)[u v - Integral v du]
= (1/5)[w sin w - integral of sin w dw]
= (1/5)[w sin w + cos w]
= (1/5) [x^5 sin(x^5) + cos(x^5)]
Check my work. The method seems correct by my algebra is often sloppy.
To evaluate the integral ∫ x^9cos(x^5)dx, we can use the technique of substitution followed by integration by parts.
Step 1: Substitution
Let u = x^5. Then, du = 5x^4 dx.
Rearranging the expression, we have dx = du/(5x^4).
Now, the integral becomes:
∫ (1/5)u^2 cos(u) du
Step 2: Integration by Parts
Integration by parts is used when you have a product of two functions to integrate. The formula for integration by parts is given by:
∫ u dv = uv - ∫ v du
In our case, we have:
u = (1/5)u^2 => du = (2/5)u du
dv = cos(u) du => v = sin(u)
Applying the formula, we get:
∫ (1/5)u^2 cos(u) du = (1/5)u^2 sin(u) - ∫ (2/5)u sin(u) du
The integral on the right-hand side can be evaluated using integration by parts once again.
Let u = u => du = du
dv = sin(u) du => v = -cos(u)
Applying the formula again, we get:
∫ (2/5)u sin(u) du = -(2/5)u cos(u) + ∫ (2/5) cos(u) du
Now, we can integrate ∫ (2/5) cos(u) du, which is simpler. The integral gives:
(2/5) sin(u)
Substituting back for u, we have:
∫ (2/5) cos(u) du = (2/5) sin(u) = (2/5) sin(x^5)
Finally, substituting all the values back into the original integral, we obtain:
∫ x^9 cos(x^5)dx = (1/5)x^2 sin(x^5) - (2/5) x^5 cos(x^5) + C
where C is the constant of integration.
So, the solution to the integral is (1/5)x^2 sin(x^5) - (2/5) x^5 cos(x^5) + C.