Evaluate `int(e^(sinx))/(cos^2x)(xcos^3x-sinx) dx`

Answer

To evaluate the integral

\[ I = \int \frac{e^{\sin x}}{\cos^2 x} \left( x \cos^3 x - \sin x \right) \, dx, \]

we will follow the steps outlined in the video transcript.

### Step 1: Simplify the Integral
We start by rewriting the integral:

\[ I = \int \frac{e^{\sin x}}{\cos^2 x} \left( x \cos^3 x - \sin x \right) \, dx. \]

Distributing \(\frac{1}{\cos^2 x}\) inside the parentheses gives:

\[ I = \int e^{\sin x} \left( x \cos x - \frac{\sin x}{\cos^2 x} \right) \, dx. \]

### Step 2: Substitution
Let \( t = \sin x \). Then, we have:

\[ dt = \cos x \, dx \quad \Rightarrow \quad dx = \frac{dt}{\cos x}. \]

Also, since \(\cos^2 x = 1 - \sin^2 x = 1 - t^2\), we can express \(\cos x\) as \(\sqrt{1 - t^2}\).

### Step 3: Change of Variables
Now, substituting these into the integral:

\[ I = \int e^t \left( \sin^{-1}(t) \cdot \sqrt{1 - t^2} - \frac{t}{1 - t^2} \right) \frac{dt}{\sqrt{1 - t^2}}. \]

This simplifies to:

\[ I = \int e^t \left( \sin^{-1}(t) - \frac{t}{(1 - t^2)^{3/2}} \right) dt. \]

### Step 4: Separate the Integral
We can separate the integral into two parts:

\[ I = \int e^t \sin^{-1}(t) \, dt - \int e^t \frac{t}{(1 - t^2)^{3/2}} \, dt. \]

### Step 5: Use Integration by Parts
For the first integral, we can apply integration by parts. Let \( u = \sin^{-1}(t) \) and \( dv = e^t dt \). Then, \( du = \frac{1}{\sqrt{1 - t^2}} dt \) and \( v = e^t \).

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du, \]

we get:

\[ \int e^t \sin^{-1}(t) \, dt = e^t \sin^{-1}(t) - \int e^t \frac{1}{\sqrt{1 - t^2}} \, dt. \]

### Step 6: Evaluate the Remaining Integral
The second integral can be evaluated using the known formula:

\[ \int e^t \frac{1}{\sqrt{1 - t^2}} \, dt = e^t \cdot \frac{1}{\sqrt{1 - t^2}} + C. \]

### Step 7: Combine Results
Combining these results, we have:

\[ I = e^t \sin^{-1}(t) - e^t \cdot \frac{1}{\sqrt{1 - t^2}} + C. \]

### Step 8: Substitute Back
Finally, substituting \( t = \sin x \) back into the equation gives:

\[ I = e^{\sin x} \sin^{-1}(\sin x) - e^{\sin x} \cdot \frac{1}{\cos x} + C. \]

Since \(\sin^{-1}(\sin x) = x\), we can write:

\[ I = e^{\sin x} \left( x - \frac{1}{\cos x} \right) + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ \int \frac{e^{\sin x}}{\cos^2 x} \left( x \cos^3 x - \sin x \right) \, dx = e^{\sin x} \left( x - \sec x \right) + C. \]