Evaluate : `intx^(2n-1)sin(x^(n))dx`

Answer

To evaluate the integral \( I = \int x^{2n-1} \sin(x^n) \, dx \), we will use a substitution method and integration by parts. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = x^n \). Then, differentiating both sides gives:
\[ dt = n x^{n-1} \, dx \quad \Rightarrow \quad dx = \frac{dt}{n x^{n-1}} \]
Since \( x = t^{1/n} \), we can express \( x^{n-1} \) in terms of \( t \):
\[ x^{n-1} = (t^{1/n})^{n-1} = t^{(n-1)/n} \]
Thus, we have:
\[ dx = \frac{dt}{n t^{(n-1)/n}} \]

### Step 2: Rewrite the Integral
Now substitute \( x^{2n-1} \) and \( dx \) into the integral:
\[ I = \int (t^{(1/n)})^{2n-1} \sin(t) \cdot \frac{dt}{n t^{(n-1)/n}} = \frac{1}{n} \int t^{(2n-1)/n} \sin(t) \cdot t^{-(n-1)/n} \, dt \]
This simplifies to:
\[ I = \frac{1}{n} \int t^{n} \sin(t) \, dt \]

### Step 3: Integration by Parts
To evaluate \( \int t^n \sin(t) \, dt \), we will use integration by parts. Let:
- \( u = t^n \) and \( dv = \sin(t) \, dt \)
Then:
- \( du = n t^{n-1} \, dt \)
- \( v = -\cos(t) \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Thus,
\[ \int t^n \sin(t) \, dt = -t^n \cos(t) - \int -\cos(t) \cdot n t^{n-1} \, dt \]
This gives:
\[ \int t^n \sin(t) \, dt = -t^n \cos(t) + n \int t^{n-1} \cos(t) \, dt \]

### Step 4: Substitute Back
Now substituting back into our expression for \( I \):
\[ I = \frac{1}{n} \left( -t^n \cos(t) + n \int t^{n-1} \cos(t) \, dt \right) \]
Substituting \( t = x^n \) back into the equation:
\[ I = \frac{1}{n} \left( -x^{n} \cos(x^n) + n \int x^{n-1} \cos(x^n) \, dx \right) \]

### Final Expression
Thus, the final expression for the integral \( I \) is:
\[ I = -\frac{x^n \cos(x^n)}{n} + \int x^{n-1} \cos(x^n) \, dx \]