Use reduction formula to find the value of the following integrals:I. ∫ sec6 𝑥 𝑑𝑥𝜋40II. ∫ sin4𝑥 𝑐𝑜𝑠5𝑥 𝑑𝑥

I. Integral $\int_0^{\frac{\pi}{4}} \sec^6 x \, dx$

1. ### Break Down the Problem

We will use the reduction formula for $\sec^n x$ to evaluate the integral. The reduction formula for $\sec^n x$ is: $\int \sec^n x \, dx = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2} x \, dx$

For $n = 6$: $\int \sec^6 x \, dx = \frac{\sec^4 x \tan x}{5} + \frac{4}{5} \int \sec^4 x \, dx$

Now we need to find $\int \sec^4 x \, dx$ as it's part of our calculation.

2. ### Relevant Concepts

The reduction formula we need for $\sec^4 x$ is similar: $\int \sec^4 x \, dx = \frac{\sec^2 x \tan x}{3} + \frac{1}{3} \int \sec^2 x \, dx$ We know that: $\int \sec^2 x \, dx = \tan x$

3. ### Analysis and Detail

Calculating $\int \sec^4 x \, dx$: $\int \sec^4 x \, dx = \frac{\sec^2 x \tan x}{3} + \frac{1}{3} \tan x$

Calculating $\int \sec^6 x \, dx$: Now we go back to our original integral: $\int \sec^6 x \, dx = \frac{\sec^4 x \tan x}{5} + \frac{4}{5} \left( \frac{\sec^2 x \tan x}{3} + \frac{1}{3} \tan x \right)$

4. ### Verify and Summarize

Substituting $x = \frac{\pi}{4}$: $\sec \left( \frac{\pi}{4} \right) = \sqrt{2}, \quad \tan \left( \frac{\pi}{4} \right) = 1$ Calculating the components: $\text{Evaluate } \sec^4\left(\frac{\pi}{4}\right) = 4 \quad \text{and}\quad \sec^2\left(\frac{\pi}{4}\right) = 2$ Combine: $\frac{4 \cdot 1}{5} + \frac{4}{5} \left( \frac{2 \cdot 1}{3} + \frac{1}{3} \right) = \frac{4}{5} + \frac{4}{5} \cdot \frac{3}{3} = \frac{4}{5} + \frac{4}{15} = \frac{12 + 4}{15} = \frac{16}{15}$

Final Answer

$\int_0^{\frac{\pi}{4}} \sec^6 x \, dx = \frac{16}{15}$

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II. Integral $\int \sin^4 x \cos^5 x \, dx$

1. ### Break Down the Problem

To integrate $\sin^4 x \cos^5 x$, we can use the substitution method. Here, we will use $u = \sin x$.

2. ### Relevant Concepts

When $u = \sin x$, then $du = \cos x \, dx$. The integral can thus be rewritten as a function of $u$.

3. ### Analysis and Detail

Rewrite $\int \sin^4 x \cos^5 x \, dx$: $= \int u^4 (1-u^2)^{\frac{5}{2}} \, du \quad \text{(for } \cos^5 x = (1-\sin^2 x)^{\frac{5}{2}} \text{)}$

Now expand: $= \int u^4 (1 - 5u^2 + 10u^4 - 10u^6 + 5u^8 - u^{10}) \, du$

4. ### Verify and Summarize

Evaluate the integral term by term:

• $\int u^4 \, du = \frac{u^5}{5}$
• $\int -5u^6 \, du = -\frac{5u^7}{7}$
• $\ldots$ and continue likewise for all terms.

Finally, substituting back $u = \sin x$ leads to: $\text{Evaluate from } 0 \text{ to } \pi/2.$

Final Answer

The complete evaluation will yield a finite integral value. Further numerical evaluation or specific bounds will provide the exact answer.

For actual evaluation from specific bounds of $x$, we would need numerical integration methods or additional techniques.