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

Solve the math by ” Wallie’s theorem ” with e

2 thoughts on “[tex]\displaystyle \int_0^{ \frac{\pi}{2} } cos^5 x\ dx = \:?\: [/tex]<br /><br />Solve the math by ” Wallie’s theorem ” with e”

1. Solution!!

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

   We have to solve this using Wallis’ theorem. As we can observe, the power is 5 which is an odd number. So, in this case,

   [tex]\displaystyle \int_{0}^{\frac{\pi}{2}}\sin ^{n}x\ dx =\displaystyle \int_{0}^{\frac{\pi}{2}}\cos ^{n}x\ dx =\dfrac{(n-1)(n-3)}{(n-0)(n-2)(n-4)}[/tex]

   Here, n is the odd number which in this case is 5. Let’s solve!

   [tex]=\dfrac{(5-1)(5-3)}{(5-0)(5-2)(5-4)}[/tex]

   [tex]=\dfrac{(4\times 2}{5\times 3\times 1}[/tex]

   [tex]=\dfrac{8}{15}[/tex]

   Another method!

   [tex]\to \displaystyle \int_0^{ \frac{\pi}{2} } cos^5 x\ dx[/tex]

   To evaluate the definite integral, first evaluate the indefinite integral.

   [tex]\to \int \cos ^{5}x\ dx[/tex]

   Write the expression as a product with the factor [tex]\cos ^{4}x[/tex].

   [tex]\to \int \cos ^{4}x\cos x\ dx[/tex]

   Use the substitution [tex]t=\sin x[/tex] to transform the integral.

   [tex]\to \int 1-2t^{2}+t^{4}\ dt[/tex]

   Use the property of integral [tex]\int f(x)\pm g(x)dx =\int f(x)dx\pm \int g(x)dx[/tex].

   [tex]\to \int 1dt-\int 2t^{2}dt+\int t^{4}dt[/tex]

   Use [tex]\int 1dx=x[/tex] to evaluate the integral.

   [tex]\to t-\int 2t^{2}dt+\int t^{4}dt[/tex]

   Evaluate the indefinite integral.

   [tex]\to t-\dfrac{2t^{3}}{3}+\int t^{4}dt[/tex]

   Use [tex]\int x^{n}dx=\dfrac{x^{n+1}}{n+1},n\neq -1[/tex] to evaluate the integral.

   [tex]\to t-\dfrac{2t^{3}}{3}+\dfrac{t^{5}}{5}[/tex]

   Substitute back [tex]t=\sin x[/tex].

   [tex]\to \sin x-\dfrac{2\sin ^{3}x}{3}+\dfrac{\sin ^{5}x}{5}[/tex]

   To evaluate the definite integral, return the limits of integration.

   [tex]\to \left. \left(\sin x-\dfrac{2\sin ^{3}x}{3}+\dfrac{\sin ^{5}x}{5}\right)\right|_{0}^{\frac{\pi }{2}}[/tex]

   Calculate the expression using [tex]\left. F(x)\right|_{a}^{b}=F(b)-F(a)[/tex].

   [tex]\to \sin \dfrac{\pi}{2}-\dfrac{2\sin ^{3}\left(\dfrac{\pi}{2}\right)}{3}+\dfrac{\sin ^{5}\left(\dfrac{\pi}{2}\right)}{5}-\left(\sin 0-\dfrac{2\sin ^{3}0}{3}+\dfrac{\sin ^{5}0}{5}\right)[/tex]

   Calculate the expression.

   [tex]\to 1-\dfrac{2\times 1^{3}}{3}+\dfrac{1^{5}}{5}-\left(0-\dfrac{2\times 0^{3}}{3}+\dfrac{0^{5}}{5}\right)[/tex]

   Simplify the expression.

   [tex]\to 1-\dfrac{2\times 1}{3}+\dfrac{1}{5}-\left(0-\dfrac{2\times 0}{3}+\dfrac{0}{5}\right)[/tex]

   [tex]\to 1-\dfrac{2}{3}+\dfrac{1}{5}-\left(-\dfrac{0}{3}+0\right)[/tex]

   [tex]\to 1-\dfrac{2}{3}+\dfrac{1}{5}-(-0)[/tex]

   [tex]\to 1-\dfrac{2}{3}+\dfrac{1}{5}+0[/tex]

   [tex]\to \dfrac{8}{15}[/tex]

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2. Step-by-step explanation:

   Given:

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

   To find: Definite integral using “Wallie’s theorem”

   Solution:

   Tip: Wallie’s theorem for odd power

   [tex] \boxed{\int_0^{ \frac{\pi}{2} } cos^nx\ dx = \int_0^{ \frac{\pi}{2} } sin^nx\ dx = \frac{2.4.6….(n – 1)}{1.3.5….n} }[/tex]

   Here,

   n is odd and it’s value is 5.

   Put the value of n in the formula of Wallie’s theorem. Numerator has only two terms ,because (5-1)=4

   and denominator have to write upto 3 terms 1.3.5 only.

   [tex]\int_0^{ \frac{\pi}{2} } cos^5x\ dx = \frac{2.(5 – 1)}{1.3.5} \\ \\ \int_0^{ \frac{\pi}{2} } cos^5x\ dx = \frac{2.4}{1.3.5} \\ \\ \int_0^{ \frac{\pi}{2} } cos^5x\ dx = \frac{8}{15} \\ \\ [/tex]

   Final Answer:

   [tex] \boxed{ \bold{\int_0^{ \frac{\pi}{2} } cos^5x\ dx = \frac{8}{15} }}\\ \\ [/tex]

   Hope it helps you.

   To learn more on brainly:

   integration of e^x(1-sinx)/(1-cosx)

   https://brainly.in/question/9459283

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