Question

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\displaystyle \int_0^{ \frac{\pi}{2} } cos^5 x\ dx = \:?\:

Solve the math by " Wallie's theorem " with explanation.

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Answer 1

Integration

We are asked to find the exact value of the following integration using Wallis theorem:

$\implies \displaystyle \int_0^{\frac{\pi}{2}} \cos^5 x \; dx$

We can see that the power is 5 which is an odd number. So, in this case we will use the following formula of Wallis theorem:

$\boxed{\displaystyle \int_0^{\frac{\pi}{2}} \cos^n x \; dx = \int_0^{\frac{\pi}{2}} \sin^n x \; dx = \dfrac{2 \cdot 4 \cdot 6 (n - 1)}{1 \cdot 3 \cdot 5 \cdot\cdot\cdot n}}$

Solution:

Substituting the known value of n in the formula of Wallis theorem, we obtain the following results:

$\implies \displaystyle \int_0^{\frac{\pi}{2}} \cos^5 x \; dx = \dfrac{2(5 - 1)}{1 \cdot 3 \cdot 5}$

$\implies \displaystyle \int_0^{\frac{\pi}{2}} \cos^5 x \; dx = \dfrac{2(4)}{1 \cdot 3 \cdot 5}$

$\implies \displaystyle \int_0^{\frac{\pi}{2}} \cos^5 x \; dx = \dfrac{8}{15}$

Therefore the required answer is:

$\boxed{\displaystyle \int_0^{\frac{\pi}{2}} \cos^5 x \; dx = \dfrac{8}{15}}$

$\rule{300}{2}$

We can also solve this problem by using another formula of Wallis theorem according to which if we have small value of n (i.e., n = 5), then the following results holds true.

$\displaystyle \int_0^{\frac{\pi}{2}} \cos^5 x \; dx = \int_0^{\frac{\pi}{2}} \sin^5 x \; dx = \dfrac{2 \cdot 4}{3 \cdot 5} = \boxed{\dfrac{8}{15}}$

$\rule{300}{2}$

The Wallis Formula

If the value of n is an even number, in that case we can use the following formula of Wallis theorem:

$\boxed{\displaystyle \int_0^{\frac{\pi}{2}} \cos^n x \; dx = \int_0^{\frac{\pi}{2}} \sin^n x \; dx = \dfrac{1 \cdot 3 \cdot 5 (n - 1)}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot n} \cdot \dfrac{\pi}{2}}$

If the value of n is an odd number, in that case we can use the following formula of Wallis theorem:

$\boxed{\displaystyle \int_0^{\frac{\pi}{2}} \cos^n x \; dx = \int_0^{\frac{\pi}{2}} \sin^n x \; dx = \dfrac{2 \cdot 4 \cdot 6 (n - 1)}{1 \cdot 3 \cdot 5 \cdot\cdot\cdot n}}$

For small vales of n, there are the following results:

$\boxed{\begin{array}{l}\displaystyle \int_0^{\frac{\pi}{2}} \cos x \; dx = \int_0^{\frac{\pi}{2}} \sin x \; dx = 1 \\ \\ \displaystyle \int_0^{\frac{\pi}{2}} \cos^2 x \; dx = \int_0^{\frac{\pi}{2}} \sin^2 x \; dx = \dfrac{1}{2} \cdot \dfrac{\pi}{2} = \dfrac{\pi}{4} \\ \\ \displaystyle \int_0^{\frac{\pi}{2}} \cos^3 x \; dx = \int_0^{\frac{\pi}{2}} \sin^3 x \; dx = \dfrac{2}{3} \\ \\ \displaystyle \int_0^{\frac{\pi}{2}} \cos^4 x \; dx = \int_0^{\frac{\pi}{2}} \sin^4 x \; dx = \dfrac{1 \cdot 3}{2 \cdot 4} \cdot \dfrac{\pi}{2} = \dfrac{3\pi}{16} \\ \\ \displaystyle \int_0^{\frac{\pi}{2}} \cos^5 x \; dx = \int_0^{\frac{\pi}{2}} \sin^5 x \; dx = \dfrac{2 \cdot 4}{4 \cdot 5} = \dfrac{8}{15} \\ \\ \displaystyle \int_0^{\frac{\pi}{2}} \cos^6 x \; dx = \int_0^{\frac{\pi}{2}} \sin^6 x \; dx = \dfrac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \cdot \dfrac{\pi}{2} = \dfrac{5\pi}{32} \end{array}}$