integrate upper limit pi/2 lower limt 0 cos^5x dx?

Question

please show steps im totally lost

integrate upper limit pi/2 lower limt 0 cos^5 x dx

Answer 1

We have cos^5(x) = cos^4(x) cos(x) = (1- sin^2(x))^2 cos(x) = (1 - 2sin^2x + sin^4(x)(cos(x)= cos(x) - 2sin^2(x) cos(x) + sin^4(x)) cos(x)

Observe that we have powers of sin(x) and the derivative of sin(x) is cos(x). So, we have a sum of functions of the type u^n du/dx, whose integral is u^(n+1)(n +1). It follows that

Integral (cos^5x dx) = sin(x) - 2/3 sin^3(x) + 1/5 sin^5(x)

If x = 0, sin(x) and it's powers are 0 and this expression gives 0.
If x = pi/2, sin(pi/2) =1, all the powers are 1 and the expression gives 1 - 2/3 + 1/5 = 8/15

Therefore, our integral is 8/15 - 0 = 8/15

Answer 2

cos^5 x dx = cos^4 x cosx dx

u = sin x
du = cos x dx

cos^4 = (cos^2 x)^2 = (1-sin^2x)^2 = (1-u^2)^2 = 1- 2u^2 + u^4

So, int [0 to pi/2] cos^5 x = int [a to b] (1-2u^2 + u^4)du

You only have to find a and b

a = sin 0 = 0

b = sin (pi/2) = 1

int (1 - 2u^2 + u^4)du = u - (2/3)u^3 + u^5 /5 + C

int (1 - 2u^2 + u^4)du = 1 - 2/3 + 1/5 - 0 = 1/3 + 1/5 = 8/15

Answer 3

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