` int _(0)^(pi//4) sec^(7) theta sin^(3) theta d theta ` is equal to

Options

A. `(1)/(12)`
B. `(3)/(12)`
C. `(5)/(12)`
D. None of these

Explanation

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{4}} \sec^7 \theta \sin^3 \theta \, d\theta, \]

we can follow these steps:

### Step 1: Rewrite the integral
We know that \(\sec \theta = \frac{1}{\cos \theta}\), so we can rewrite \(\sec^7 \theta\) as \(\frac{1}{\cos^7 \theta}\). Therefore, we have:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin^3 \theta}{\cos^7 \theta} \, d\theta. \]

### Step 2: Express \(\sin^3 \theta\) in terms of \(\tan \theta\)
Using the identity \(\sin \theta = \tan \theta \cos \theta\), we can express \(\sin^3 \theta\) as:

\[ \sin^3 \theta = \tan^3 \theta \cos^3 \theta. \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{4}} \tan^3 \theta \cdot \frac{\cos^3 \theta}{\cos^7 \theta} \, d\theta = \int_{0}^{\frac{\pi}{4}} \tan^3 \theta \sec^4 \theta \, d\theta. \]

### Step 3: Use substitution
Let \(t = \tan \theta\). Then, the derivative \(dt = \sec^2 \theta \, d\theta\) implies \(d\theta = \frac{dt}{\sec^2 \theta}\).

When \(\theta = 0\), \(t = 0\) and when \(\theta = \frac{\pi}{4}\), \(t = 1\). Therefore, we can change the limits of integration:

\[ I = \int_{0}^{1} t^3 \sec^4(\tan^{-1} t) \cdot \frac{dt}{\sec^2(\tan^{-1} t)}. \]

### Step 4: Simplify the integral
Using the identity \(\sec^2(\tan^{-1} t) = 1 + t^2\), we have:

\[ \sec^4(\tan^{-1} t) = (1 + t^2)^2. \]

Thus, the integral becomes:

\[ I = \int_{0}^{1} t^3 (1 + t^2)^2 \, dt. \]

### Step 5: Expand and integrate
Expanding \((1 + t^2)^2\):

\[ (1 + t^2)^2 = 1 + 2t^2 + t^4. \]

So, we have:

\[ I = \int_{0}^{1} t^3 (1 + 2t^2 + t^4) \, dt = \int_{0}^{1} (t^3 + 2t^5 + t^7) \, dt. \]

Now, we can integrate term by term:

\[ I = \left[ \frac{t^4}{4} + \frac{2t^6}{6} + \frac{t^8}{8} \right]_{0}^{1} = \frac{1}{4} + \frac{1}{3} + \frac{1}{8}. \]

### Step 6: Find a common denominator and sum
The common denominator for \(4\), \(3\), and \(8\) is \(24\):

\[ I = \frac{6}{24} + \frac{8}{24} + \frac{3}{24} = \frac{17}{24}. \]

Thus, the final answer is:

\[ I = \frac{17}{24}. \]