Question

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

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

Answer : $\frac{8}{15}$


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Answer 1

Step-by-step explanation:

Given:

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

To find: Definite integral using "Wallie's theorem"

Solution:

Tip: Wallie's theorem for odd power

$\boxed{\int_0^{ \frac{\pi}{2} } cos^nx\ dx = \int_0^{ \frac{\pi}{2} } sin^nx\ dx = \frac{2.4.6....(n - 1)}{1.3.5....n} }$

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.

$\int_0^{ \frac{\pi}{2} } cos^5x\ dx = \frac{2.(5 - 1)}{1.3.5} \\ \\ \int_0^{ \frac{\pi}{2} } cos^5x\ dx = \frac{2.4}{1.3.5} \\ \\ \int_0^{ \frac{\pi}{2} } cos^5x\ dx = \frac{8}{15}$

Final Answer:

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

Hope it helps you.

To learn more on brainly:

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

https://brainly.in/question/9459283

Answer 2

Solution!!

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

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,

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

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

$=\dfrac{(5-1)(5-3)}{(5-0)(5-2)(5-4)}$

$=\dfrac{(4\times 2}{5\times 3\times 1}$

$=\dfrac{8}{15}$

Another method!

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

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

$\to \int \cos ^{5}x\ dx$

Write the expression as a product with the factor $\cos ^{4}x$.

$\to \int \cos ^{4}x\cos x\ dx$

Use the substitution $t=\sin x$ to transform the integral.

$\to \int 1-2t^{2}+t^{4}\ dt$

Use the property of integral $\int f(x)\pm g(x)dx =\int f(x)dx\pm \int g(x)dx$.

$\to \int 1dt-\int 2t^{2}dt+\int t^{4}dt$

Use $\int 1dx=x$ to evaluate the integral.

$\to t-\int 2t^{2}dt+\int t^{4}dt$

Evaluate the indefinite integral.

$\to t-\dfrac{2t^{3}}{3}+\int t^{4}dt$

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

$\to t-\dfrac{2t^{3}}{3}+\dfrac{t^{5}}{5}$

Substitute back $t=\sin x$.

$\to \sin x-\dfrac{2\sin ^{3}x}{3}+\dfrac{\sin ^{5}x}{5}$

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

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

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

$\to \sin \dfrac{\pi}{2}-\dfrac{2\sin ^{3}\left(\dfrac{\pi}{2}\right)}{3}+\dfrac{\sin ^{5}\left(\dfrac{\pi}{2}\right)}{5}-\left(\sin 0-\dfrac{2\sin ^{3}0}{3}+\dfrac{\sin ^{5}0}{5}\right)$

Calculate the expression.

$\to 1-\dfrac{2\times 1^{3}}{3}+\dfrac{1^{5}}{5}-\left(0-\dfrac{2\times 0^{3}}{3}+\dfrac{0^{5}}{5}\right)$

Simplify the expression.

$\to 1-\dfrac{2\times 1}{3}+\dfrac{1}{5}-\left(0-\dfrac{2\times 0}{3}+\dfrac{0}{5}\right)$

$\to 1-\dfrac{2}{3}+\dfrac{1}{5}-\left(-\dfrac{0}{3}+0\right)$

$\to 1-\dfrac{2}{3}+\dfrac{1}{5}-(-0)$

$\to 1-\dfrac{2}{3}+\dfrac{1}{5}+0$

$\to \dfrac{8}{15}$