Question

Sec^8/cosecx integration

Answer 1

$\frac{1}{7} \sec^7 x + C$

Answer 2

To integrate the given expression, we first rewrite it in terms of $\sin x$ and $\cos x$.

The given expression is $\frac{\sec^8 x}{\csc x}$. We know that $\sec x = \frac{1}{\cos x}$ and $\csc x = \frac{1}{\sin x}$.

Substitute these into the expression: $\frac{\sec^8 x}{\csc x} = \frac{\left(\frac{1}{\cos x}\right)^8}{\frac{1}{\sin x}} = \frac{\frac{1}{\cos^8 x}}{\frac{1}{\sin x}} = \frac{1}{\cos^8 x} \cdot \sin x = \frac{\sin x}{\cos^8 x}$

Now, we need to find the integral of $\frac{\sin x}{\cos^8 x}$: $I = \int \frac{\sin x}{\cos^8 x} dx$

We can solve this integral using the substitution method. Let $u = \cos x$. Differentiating both sides with respect to $x$: $\frac{du}{dx} = -\sin x$ $du = -\sin x \, dx$ So, $\sin x \, dx = -du$.

Substitute $u$ and $du$ into the integral: $I = \int \frac{-du}{u^8} = -\int u^{-8} du$

Now, integrate $u^{-8}$ using the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$: $I = -\left( \frac{u^{-8+1}}{-8+1} \right) + C$ $I = -\left( \frac{u^{-7}}{-7} \right) + C$ $I = \frac{1}{7} u^{-7} + C$

Finally, substitute back $u = \cos x$: $I = \frac{1}{7} (\cos x)^{-7} + C$ $I = \frac{1}{7 \cos^7 x} + C$ Since $\frac{1}{\cos x} = \sec x$, we can write: $I = \frac{1}{7} \sec^7 x + C$

The final answer is $\boxed{\frac{1}{7} \sec^7 x + C}$.

Explanation of the solution:

1. Rewrite the integrand $\frac{\sec^8 x}{\csc x}$ as $\frac{\sin x}{\cos^8 x}$ using $\sec x = 1/\cos x$ and $\csc x = 1/\sin x$.
2. Use substitution: Let $u = \cos x$, so $du = -\sin x \, dx$.
3. Substitute into the integral: $\int \frac{-du}{u^8} = -\int u^{-8} du$.
4. Integrate using the power rule: $-\frac{u^{-7}}{-7} + C = \frac{1}{7} u^{-7} + C$.
5. Substitute back $u = \cos x$ to get $\frac{1}{7} (\cos x)^{-7} + C = \frac{1}{7} \sec^7 x + C$.

Answer:

The integration of $\frac{\sec^8 x}{\csc x}$ is $\frac{1}{7} \sec^7 x + C$.