Using Integration by parts, show that In = (pi)/(2) ^n-2 x x + ((n-1))/(n) In-2. Hence evaluate integral0^(pi)/(2) 5 x dx.

The problem asks us to first show a reduction formula for the integral of $\cos^n x$ using integration by parts, and then use it to evaluate a definite integral.

Part 1: Show that $nI_n = \sin x \cos^{n-1} x + (n-1)I_{n-2}$ for $I_n = \int \cos^n x \, dx$.

Step 1: Set up the integral for integration by parts.
Let $I_n = \int \cos^n x \, dx = \int \cos^{n-1} x \cdot \cos x \, dx$.
We use integration by parts formula: $\int u \, dv = uv - \int v \, du$.
Let $u = \cos^{n-1} x$ and $dv = \cos x \, dx$.

Step 2: Find $du$ and $v$.
Differentiating $u$: $du = (n-1)\cos^{n-2} x (-\sin x) \, dx = -(n-1)\cos^{n-2} x \sin x \, dx$.
Integrating $dv$: $v = \int \cos x \, dx = \sin x$.

Step 3: Apply the integration by parts formula.

$$I_n = \cos^{n-1} x \sin x - \int \sin x \cdot (-(n-1)\cos^{n-2} x \sin x) \, dx$$ $$I_n = \cos^{n-1} x \sin x + (n-1) \int \sin^2 x \cos^{n-2} x \, dx$$

Step 4: Use the identity $\sin^2 x = 1 - \cos^2 x$.

$$I_n = \cos^{n-1} x \sin x + (n-1) \int (1 - \cos^2 x) \cos^{n-2} x \, dx$$ $$I_n = \cos^{n-1} x \sin x + (n-1) \int (\cos^{n-2} x - \cos^n x) \, dx$$ $$I_n = \cos^{n-1} x \sin x + (n-1) \int \cos^{n-2} x \, dx - (n-1) \int \cos^n x \, dx$$

Step 5: Substitute $I_{n-2}$ and $I_n$ back into the equation.

$$I_n = \cos^{n-1} x \sin x + (n-1)I_{n-2} - (n-1)I_n$$

Step 6: Rearrange the equation to solve for $I_n$.

$$I_n + (n-1)I_n = \cos^{n-1} x \sin x + (n-1)I_{n-2}$$ $$I_n (1 + n - 1) = \cos^{n-1} x \sin x + (n-1)I_{n-2}$$ $$nI_n = \cos^{n-1} x \sin x + (n-1)I_{n-2}$$

This proves the reduction formula.

Part 2: Hence evaluate $\int_0^{\frac{\pi}{2}} \cos^5 x \, dx$.

Step 1: Apply the reduction formula to the definite integral.
Let $J_n = \int_0^{\frac{\pi}{2}} \cos^n x \, dx$.
Using the reduction formula from Part 1:

$$n J_n = \left[ \sin x \cos^{n-1} x \right]_0^{\frac{\pi}{2}} + (n-1) J_{n-2}$$

Step 2: Evaluate the boundary term.
For $n > 1$:

$$\left[ \sin x \cos^{n-1} x \right]_0^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) \cos^{n-1}\left(\frac{\pi}{2}\right) - \sin(0) \cos^{n-1}(0)$$ $$= (1) \cdot (0) - (0) \cdot (1) = 0$$

So, for $n > 1$, the reduction formula for definite integrals from 0 to $\frac{\pi}{2}$ simplifies to:

$$n J_n = (n-1) J_{n-2} \implies J_n = \frac{n-1}{n} J_{n-2}$$

Step 3: Use the reduction formula to evaluate $J_5$.

$$J_5 = \frac{5-1}{5} J_{5-2} = \frac{4}{5} J_3$$ $$J_3 = \frac{3-1}{3} J_{3-2} = \frac{2}{3} J_1$$

Step 4: Calculate $J_1$.

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