The value of `int_(-1)^(1)(sin^(-1)x+(x^(5)+x^(3)-1)/(cosx))dx` is equal to

Options

A. `tan1`
B. 0
C. `2tan1`
D. `-2tan1`

Answer

To solve the integral

\[ I = \int_{-1}^{1} \left( \sin^{-1}x + \frac{x^5 + x^3 - 1}{\cos x} \right) dx, \]

we can break it down into three separate integrals:

\[ I = \int_{-1}^{1} \sin^{-1}x \, dx + \int_{-1}^{1} \frac{x^5}{\cos x} \, dx + \int_{-1}^{1} \frac{x^3}{\cos x} \, dx - \int_{-1}^{1} \frac{1}{\cos x} \, dx. \]

### Step 1: Evaluate \(\int_{-1}^{1} \sin^{-1}x \, dx\)

The function \(\sin^{-1}x\) is an odd function because:

\[ \sin^{-1}(-x) = -\sin^{-1}(x). \]

Thus, the integral of an odd function over a symmetric interval around zero is zero:

\[ \int_{-1}^{1} \sin^{-1}x \, dx = 0. \]

### Step 2: Evaluate \(\int_{-1}^{1} \frac{x^5}{\cos x} \, dx\)

Next, we check the function \(\frac{x^5}{\cos x}\):

\[ \frac{-x^5}{\cos(-x)} = \frac{-x^5}{\cos x}. \]

This shows that \(\frac{x^5}{\cos x}\) is also an odd function. Therefore, its integral over the symmetric interval is also zero:

\[ \int_{-1}^{1} \frac{x^5}{\cos x} \, dx = 0. \]

### Step 3: Evaluate \(\int_{-1}^{1} \frac{x^3}{\cos x} \, dx\)

Similarly, for \(\frac{x^3}{\cos x}\):

\[ \frac{-x^3}{\cos(-x)} = \frac{-x^3}{\cos x}. \]

This indicates that \(\frac{x^3}{\cos x}\) is an odd function as well, leading to:

\[ \int_{-1}^{1} \frac{x^3}{\cos x} \, dx = 0. \]

### Step 4: Evaluate \(-\int_{-1}^{1} \frac{1}{\cos x} \, dx\)

Now we need to evaluate \(-\int_{-1}^{1} \frac{1}{\cos x} \, dx\). The function \(\frac{1}{\cos x}\) is an even function because:

\[ \frac{1}{\cos(-x)} = \frac{1}{\cos x}. \]

Thus, we can simplify the integral:

\[ -\int_{-1}^{1} \frac{1}{\cos x} \, dx = -2 \int_{0}^{1} \frac{1}{\cos x} \, dx. \]

### Step 5: Final Calculation

The integral \(\int_{0}^{1} \sec x \, dx\) can be computed. The integral of \(\sec x\) is:

\[ \int \sec x \, dx = \ln | \sec x + \tan x | + C. \]

Thus,

\[ \int_{0}^{1} \sec x \, dx = \left[ \ln | \sec x + \tan x | \right]_{0}^{1} = \ln | \sec(1) + \tan(1) | - \ln | \sec(0) + \tan(0) |. \]

Since \(\sec(0) = 1\) and \(\tan(0) = 0\):

\[ \ln | \sec(0) + \tan(0) | = \ln(1) = 0. \]

Therefore,

\[ \int_{0}^{1} \sec x \, dx = \ln | \sec(1) + \tan(1) |. \]

Finally, we have:

\[ I = -2 \ln | \sec(1) + \tan(1) |. \]

### Conclusion

The value of the integral is:

\[ I = -2 \ln | \sec(1) + \tan(1) |. \]