The value of the integral `int((1-costheta)^(2/7))/((1+costheta)^(9/7))d theta` is
(a)`7/(11)(t a ntheta/2)^((11)/7)+C`
(b) `7/(11)(costheta/2)^((11)/7)+C`
(c)`7/(11)(sintheta/2)^((11)/7)+C`
(d) none of these

Options

A. `(7)/(11)("tan"(theta)/(2))^((11)/(7))+C`
B. `(7)/(11)("cos"(theta)/(2))^((11)/(7))+C`
C. `(7)/(11)("sin"(theta)/(2))^((11)/(7))+C`
D. none of these

Explanation

To solve the integral

\[ I = \int \frac{(1 - \cos \theta)^{\frac{2}{7}}}{(1 + \cos \theta)^{\frac{9}{7}}} d\theta, \]

we can use trigonometric identities and substitutions. Let's go through the steps:

### Step 1: Rewrite the integral using trigonometric identities
We know that:

\[ 1 - \cos \theta = 2 \sin^2 \left( \frac{\theta}{2} \right) \quad \text{and} \quad 1 + \cos \theta = 2 \cos^2 \left( \frac{\theta}{2} \right). \]

Substituting these into the integral gives:

\[ I = \int \frac{(2 \sin^2 \left( \frac{\theta}{2} \right))^{\frac{2}{7}}}{(2 \cos^2 \left( \frac{\theta}{2} \right))^{\frac{9}{7}}} d\theta. \]

### Step 2: Simplify the integral
This can be simplified as follows:

\[ I = \int \frac{2^{\frac{2}{7}} \sin^{\frac{4}{7}} \left( \frac{\theta}{2} \right)}{2^{\frac{9}{7}} \cos^{\frac{18}{7}} \left( \frac{\theta}{2} \right)} d\theta. \]

Factoring out the constants gives:

\[ I = \frac{2^{\frac{2}{7} - \frac{9}{7}}}{1} \int \frac{\sin^{\frac{4}{7}} \left( \frac{\theta}{2} \right)}{\cos^{\frac{18}{7}} \left( \frac{\theta}{2} \right)} d\theta = 2^{-\frac{7}{7}} \int \frac{\sin^{\frac{4}{7}} \left( \frac{\theta}{2} \right)}{\cos^{\frac{18}{7}} \left( \frac{\theta}{2} \right)} d\theta. \]

This simplifies to:

\[ I = \frac{1}{2} \int \frac{\sin^{\frac{4}{7}} \left( \frac{\theta}{2} \right)}{\cos^{\frac{18}{7}} \left( \frac{\theta}{2} \right)} d\theta. \]

### Step 3: Substitution
Let \( t = \tan \left( \frac{\theta}{2} \right) \). Then, we have:

\[ d\theta = \frac{2}{1 + t^2} dt. \]

Also, we can express \(\sin\) and \(\cos\) in terms of \(t\):

\[ \sin \left( \frac{\theta}{2} \right) = \frac{t}{\sqrt{1 + t^2}}, \quad \cos \left( \frac{\theta}{2} \right) = \frac{1}{\sqrt{1 + t^2}}. \]

### Step 4: Substitute into the integral
Substituting these into the integral gives:

\[ I = \frac{1}{2} \int \frac{\left(\frac{t}{\sqrt{1+t^2}}\right)^{\frac{4}{7}}}{\left(\frac{1}{\sqrt{1+t^2}}\right)^{\frac{18}{7}}} \cdot \frac{2}{1+t^2} dt. \]

This simplifies to:

\[ I = \int t^{\frac{4}{7}} (1 + t^2)^{\frac{9}{7} - 1} dt. \]

### Step 5: Integrate
Now, we can integrate:

\[ I = \int t^{\frac{4}{7}} (1 + t^2)^{-\frac{2}{7}} dt. \]

Using the formula for integration, we get:

\[ I = \frac{7}{11} t^{\frac{11}{7}} + C. \]

### Step 6: Back substitute
Finally, substituting back \( t = \tan \left( \frac{\theta}{2} \right) \):

\[ I = \frac{7}{11} \left( \tan \left( \frac{\theta}{2} \right) \right)^{\frac{11}{7}} + C. \]

### Conclusion
Thus, the value of the integral is:

\[ I = \frac{7}{11} \left( \tan \left( \frac{\theta}{2} \right) \right)^{\frac{11}{7}} + C. \]