`(i)intsec^(7) x. sin x dx" "(ii) int(1)/(sinx. cos^(2) x)dx`

Answer

Let's solve the given integrals step by step.

### (i) Evaluate the integral \( I_1 = \int \sec^7 x \sin x \, dx \)

1. **Substitution**: We know that \( \sec x = \frac{1}{\cos x} \) and \( \sin x = \sqrt{1 - \cos^2 x} \). Thus, we can rewrite the integral:
\[ I_1 = \int \sec^7 x \sin x \, dx = \int \frac{\sin x}{\cos^7 x} \, dx \]

2. **Let \( t = \cos x \)**: Then, \( dt = -\sin x \, dx \) or \( \sin x \, dx = -dt \). Substituting these into the integral gives:
\[ I_1 = \int \frac{-dt}{t^7} = -\int t^{-7} \, dt \]

3. **Integrate**: Using the power rule for integration:
\[ -\int t^{-7} \, dt = -\left( \frac{t^{-6}}{-6} \right) + C = \frac{1}{6} t^{-6} + C \]

4. **Back Substitute**: Replace \( t \) back with \( \cos x \):
\[ I_1 = \frac{1}{6} \cos^{-6} x + C = \frac{1}{6 \cos^6 x} + C \]

### Final Result for (i):
\[ I_1 = \frac{1}{6 \cos^6 x} + C \]

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### (ii) Evaluate the integral \( I_2 = \int \frac{1}{\sin x \cos^2 x} \, dx \)

1. **Rewrite the Integral**: We can multiply the numerator and denominator by \( \sin x \):
\[ I_2 = \int \frac{\sin x}{\sin^2 x \cos^2 x} \, dx = \int \frac{\sin x}{(1 - \cos^2 x) \cos^2 x} \, dx \]

2. **Let \( u = \cos x \)**: Then, \( du = -\sin x \, dx \) or \( \sin x \, dx = -du \). Substitute:
\[ I_2 = \int \frac{-du}{(1 - u^2) u^2} \]

3. **Partial Fraction Decomposition**: We can express the integrand as:
\[ \frac{-1}{(1 - u^2) u^2} = \frac{A}{u} + \frac{B}{1-u} + \frac{C}{1+u} \]
Solving for \( A, B, C \) gives:
\[ I_2 = -\int \left( \frac{1}{u} + \frac{1}{1-u} + \frac{1}{1+u} \right) \, du \]

4. **Integrate**: Each term integrates to:
\[ I_2 = -\left( \ln |u| - \ln |1-u| + \ln |1+u| \right) + C \]

5. **Back Substitute**: Replace \( u \) back with \( \cos x \):
\[ I_2 = -\left( \ln |\cos x| - \ln |1 - \cos x| + \ln |1 + \cos x| \right) + C \]

### Final Result for (ii):
\[ I_2 = -\ln |\cos x| + \ln |1 - \cos x| - \ln |1 + \cos x| + C \]

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