`I= int root(3) ( 1+ 3 sin x) cos x dx`.

Answer

To solve the integral \( I = \int \sqrt[3]{1 + 3 \sin x} \cos x \, dx \), we will use substitution and the basic integration formula.

### Step 1: Substitution
Let \( t = 1 + 3 \sin x \). Then, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = 3 \cos x \implies dt = 3 \cos x \, dx \implies \cos x \, dx = \frac{1}{3} dt \]

### Step 2: Rewrite the Integral
Now, we can rewrite the integral in terms of \( t \):
\[ I = \int \sqrt[3]{t} \cdot \frac{1}{3} dt = \frac{1}{3} \int t^{1/3} \, dt \]

### Step 3: Integrate
Using the power rule for integration, where \( \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \), we have:
\[ \int t^{1/3} \, dt = \frac{t^{1/3 + 1}}{1/3 + 1} = \frac{t^{4/3}}{4/3} = \frac{3}{4} t^{4/3} \]
Thus,
\[ I = \frac{1}{3} \cdot \frac{3}{4} t^{4/3} + C = \frac{1}{4} t^{4/3} + C \]

### Step 4: Substitute Back
Now, we substitute back \( t = 1 + 3 \sin x \):
\[ I = \frac{1}{4} (1 + 3 \sin x)^{4/3} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = \frac{1}{4} (1 + 3 \sin x)^{4/3} + C \]

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