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Prove that, for even powers of sine,$$ \int_0^{\frac{\pi}{2}} \sin^{2n} x dx = \frac{1 \cdot 3 \cdot 5 \cdots \cdots (2n - 1)}{2 \cdot 4 \cdot 6 \cdots \cdots 2n} \frac{\pi}{2} $$

Hence, $[(2 n+1)+1] \int_{0}^{\frac{\pi}{2}} \sin ^{2(n+1)} x d x=(2 n+1) \int_{0}^{\frac{\pi}{2}} \sin ^{2 n} x d x=(2 n+$1) $\frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 n-1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 n} \cdot \frac{\pi}{2}=\frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 n-1) \cdot(2(n+1)-1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 n} \frac{\pi}{2}$ . Andthen we have that $\int_{0}^{\frac{\pi}{2}} \sin ^{2(n+1)} x d x=\frac{1}{2(n+1)} \frac{1 \cdot 3 \cdot 5 \cdot \ldots(2 n-1) \cdot(2(n+1)-1)}{2 n} \frac{\pi}{2}=$$\frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 n-1) \cdot(2(n+1)-1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 n \cdot 2(n+1)} \frac{\pi}{2}$Thus the formula is true for $n+1$ and then we have that it's true for every $n$ integer with $n \geq 1 .$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Missouri State University

Oregon State University

University of Nottingham

Lectures

01:11

In mathematics, integratio…

06:55

In grammar, determiners ar…

06:33

Prove that, for even power…

08:52

(a) Use the reduction form…

03:41

Use integration by parts t…

03:04

Given that $n$ is a positi…

11:18

13:58

02:32

$$\text{Show that}\int_{0}…

01:55

Evaluate $I=\int_{0}^{2 \p…

02:25

Prove the formula, where $…

01:36

Prove that $$0 \leq \int_{…

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Numerade Educator

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