2 answers

Evaluate the indefinite integral:∫sqrt(10x−x^2)dx ?

Question:

Evaluate the indefinite integral:∫sqrt(10x−x^2)dx ?

Answers

#20/3x^(3/2)-1/2x^2+c#

Explanation:

#int" "sqrt(10x-x^2)" "dx#

Complete the square,

#int" "sqrt(25-(x-5)^2)" "dx#

Substitute #u=x-5#,

#int" "sqrt(25-u^2)" "du#

Substitute #u=5sin(v)# and #du=5cos(v)#

#int" "5cos(v) sqrt(25-25sin^2(v))" "dv#

Simplify,

#int" "(5cos(v)) (5cos(v))" "dv#

Refine,

#int" "25cos^2(v)" "dv#

Take out the constant,

#25int" "cos^2(v)" "dv#

Apply double angle formulae,

#25int" "(1+cos(2v))/2" "dv#

Take out the constant,

#25/2int" "1+cos(2v)" "dv#

Integrate,

#25/2(v+1/2sin(2v))"+c#

Substitute back #v=arcsin(u/5)# and #u=x-5#

#25/2(arcsin((x-5)/5)+cancel(1/2sin)(cancel(2arcsin)((x-5)/5)))"+c#

Simplify,

#25/2(arcsin((x-5)/5))+25/2((x-5)/5)+c#

Refine,

#25/2arcsin((x-5)/5)+5/2(x-5)+c#, where #c# is the constant of integration.

Tadaa :D

.

#=1/2(((x-5)sqrt(-5(x^2-10x+20))) + 25/2arcsin((x-5)/5) +c#

Explanation:

What is #int sqrt(10x - x^2) dx# ?

Note that the domain of the function being integrated is where the inner quadratic is positive, i.e. #x in [0, 10]#

This expression can be integrated using substitutions. Though a possible pathway for integration doesn't immediately present itself, if we compete the square, then a trigonometric substitution can be carried out:

#10x - x^2 = 25 - (x-5)^2#

Which, we notice, is in the classic trigonometric substitution form, i.e. the square of a number minus the square of a linear #x# function.

First, to get rid of the linear, we let #u = x-5#, which gives #du=dx#, so we can rewrite the above integral as:

#int sqrt(25-u^2) du#

Now for the second substitution, let #u = 5sintheta#, which changes the integral to:

#int sqrt(25 - 25sin^2theta) dx#
#= int abs(5costheta) dx# (we can ignore the absolute value brackets)

Of course, the #dx# isn't helping, so we differentiate the substitution equation to get: #du = 5costheta d theta#, so the integral becomes:

#25 int cos^2 theta d theta#

Now we can use a double angle formula to make integrating #cos^2 theta# easier:

#cos(2 theta) = 2cos^2theta -1#
#:. cos^2theta = 1/2(cos(2theta)+1)#
So the integral becomes:

#25/2 int cos(2theta) + 1 d theta#
#=25/2(1/2sin(2 theta) + theta) + c#
#= 25/2(sinthetacostheta + theta) + c# (using a double-angle formula)

Now, #sintheta = u/5 = (x-5)/5#
Hence, #cos theta = sqrt(1-u^2/25) = sqrt((-x^2+10x-20)/25)#
And, #theta = arcsin(u/5) = arcsin((x-5)/5)#

#int sqrt(10x - x^2) dx#
#= 25/2(((x-5)sqrt(-5(x^2-20x+20)))/25 + arcsin((x-5)/5)) +c#
#=1/2(((x-5)sqrt(-5(x^2-10x+20))) + 25/2arcsin((x-5)/5) +c#

.

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