To evaluate the indefinite integral \(\int x^7 \sin(x^8) , dx\), we can use the method of substitution.
Let: \[ u = x^8 \] Then, the differential \(du\) is given by: \[ du = 8x^7 , dx \quad \Rightarrow \quad dx = \frac{du}{8x^7} \] Now, substitute \(u\) in the integral. The \(x^7\) in \(dx\) will cancel out: \[ \int x^7 \sin(x^8) , dx = \int x^7 \sin(u) \cdot \frac{du}{8x^7} = \frac{1}{8} \int \sin(u) , du \]
The integral of \(\sin(u)\) is: \[ \int \sin(u) , du = -\cos(u) \]
Thus, we have: \[ \frac{1}{8} \int \sin(u) , du = \frac{1}{8} (-\cos(u)) = -\frac{1}{8} \cos(u) + C \]
Now substitute back \(u = x^8\): \[ -\frac{1}{8} \cos(x^8) + C \]
So, the final answer is: \[ \int x^7 \sin(x^8) , dx = -\frac{1}{8} \cos(x^8) + C \]