The value of `int_a^b (x^7+sinx)/(cos x) dx` where a+b=0 is

Options

A. `2b-a sin (b-a)`
B. `a+3b cos (b-a)`
C. `sina-(b-a) cos b`
D. 0

Answer

To solve the integral \( \int_a^b \frac{x^7 + \sin x}{\cos x} \, dx \) given that \( a + b = 0 \), we can follow these steps:

### Step 1: Understand the limits of integration
Since \( a + b = 0 \), we can express \( b \) as \( b = -a \). This means our integral can be rewritten as:
\[ \int_a^{-a} \frac{x^7 + \sin x}{\cos x} \, dx \]

### Step 2: Use the property of definite integrals
We can use the property of definite integrals that states:
\[ \int_a^{-a} f(x) \, dx = -\int_{-a}^a f(-x) \, dx \]
Thus, we can rewrite our integral as:
\[ \int_a^{-a} \frac{x^7 + \sin x}{\cos x} \, dx = -\int_{-a}^a \frac{(-x)^7 + \sin(-x)}{\cos(-x)} \, dx \]

### Step 3: Simplify the integrand
Now we simplify the integrand:
- \( (-x)^7 = -x^7 \) (since 7 is odd)
- \( \sin(-x) = -\sin x \) (since sine is an odd function)
- \( \cos(-x) = \cos x \) (since cosine is an even function)

Putting this together, we have:
\[ \int_a^{-a} \frac{x^7 + \sin x}{\cos x} \, dx = -\int_{-a}^a \frac{-x^7 - \sin x}{\cos x} \, dx = \int_{-a}^a \frac{x^7 + \sin x}{\cos x} \, dx \]

### Step 4: Combine the results
Now we have:
\[ \int_a^{-a} \frac{x^7 + \sin x}{\cos x} \, dx = -\int_a^{-a} \frac{x^7 + \sin x}{\cos x} \, dx \]

Let \( I = \int_a^{-a} \frac{x^7 + \sin x}{\cos x} \, dx \). Then we can write:
\[ I = -I \]

### Step 5: Solve for \( I \)
Adding \( I \) to both sides gives:
\[ 2I = 0 \implies I = 0 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_a^b \frac{x^7 + \sin x}{\cos x} \, dx = 0 \]

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