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Math  /  Calculus

Question∫s4/3sin⁡(s7/3−3)ds∫s4/3sin⁡(s7/3−3)ds=\begin{array}{l}\int s^{4 / 3} \sin \left(s^{7 / 3}-3\right) d s \\ \int s^{4 / 3} \sin \left(s^{7 / 3}-3\right) d s=\end{array}

Studdy Solution

STEP 1

1. The integral is in the form ∫s4/3sin⁡(s7/3−3) ds\int s^{4/3} \sin(s^{7/3} - 3) \, ds.
2. We will use substitution to simplify the integration process.

STEP 2

1. Identify a substitution to simplify the integral.
2. Perform the substitution.
3. Integrate the simplified expression.
4. Substitute back to the original variable.

STEP 3

Identify a substitution that will simplify the integral. Notice that the derivative of s7/3s^{7/3} is related to s4/3s^{4/3}.
Let u=s7/3−3 u = s^{7/3} - 3 .
Then, differentiate u u with respect to s s :
duds=73s4/3 \frac{du}{ds} = \frac{7}{3} s^{4/3}
Rearrange to express ds ds in terms of du du :
ds=37s−4/3 du ds = \frac{3}{7} s^{-4/3} \, du

STEP 4

Substitute u=s7/3−3 u = s^{7/3} - 3 and ds=37s−4/3 du ds = \frac{3}{7} s^{-4/3} \, du into the integral:
∫s4/3sin⁡(s7/3−3) ds=∫s4/3sin⁡(u)⋅37s−4/3 du \int s^{4/3} \sin(s^{7/3} - 3) \, ds = \int s^{4/3} \sin(u) \cdot \frac{3}{7} s^{-4/3} \, du
Simplify the expression:
=37∫sin⁡(u) du = \frac{3}{7} \int \sin(u) \, du

STEP 5

Integrate the simplified expression:
∫sin⁡(u) du=−cos⁡(u)+C \int \sin(u) \, du = -\cos(u) + C
Thus, the integral becomes:
37(−cos⁡(u)+C) \frac{3}{7} \left(-\cos(u) + C\right)
=−37cos⁡(u)+C = -\frac{3}{7} \cos(u) + C

STEP 6

Substitute back the original variable u=s7/3−3 u = s^{7/3} - 3 :
−37cos⁡(s7/3−3)+C -\frac{3}{7} \cos(s^{7/3} - 3) + C
The solution to the integral is:
−37cos⁡(s7/3−3)+C \boxed{-\frac{3}{7} \cos(s^{7/3} - 3) + C}

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