How do I solve the following using trig substitutions? (x^2)/((25-x^2)^3/2)

1 Answer
Jul 3, 2018

#" "#
#color(blue)(x/[sqrt(25-x^2)]-sin^(-1)(x/5)+C#

Explanation:

#" "#
I am assuming that your question is:

#color(red)(int (x^2)/[25-x^2]^(3/2) " "dx#

Rewrite the problem as follows:

#color(red)(int (x^2)/[[sqrt[25-x^2]]^(3) " "dx# #color(blue)(Exp.1#

Use Trigonometric SIN substitution

Let #color(green)(sqrt(a^2-x^2) rArr x = a sin theta#

Consider the denominator part [DR] of our problem:

#[sqrt[25-x^2]]^(3)#

#rArr [sqrt[[5^2-x^2]]]^3#

Using sin substitution, substitute #color(red)(x=5 sin theta# to obtain

#rArr [sqrt[[25-(5 sin theta)^2]]]^3#

#rArr [sqrt[[25-(5^2 sin^2 theta)]]]^3#

#rArr [sqrt[[25-25 sin^2 theta]]]^3#

#rArr [sqrt[[25(1-sin^2 theta)]]^3#

We use the following trigonometric relationship to simplify further:

#color(green)(sin^2 theta + cos^2 theta = 1#

#color(green)(cos^2 theta = 1 - sin^2 theta)#

#rArr [sqrt[25(cos^2 theta)]]^3#

The above expression reduces to

#color(red)([5 |cos theta|]^3)#

We consider #color(blue)(cos theta)# is positive

#rArr [ 5 cos theta ]^3#

#rArr 5^3 cos^3 theta#

#rArr 125 cos^3 theta#

We already have

#color(blue)(x=5 sin theta)#

Using the above relationship we obtain

#color(blue)(dx = 5 cos theta" "d theta#

Next, we will rewrite #color(blue)(Exp.1# as

#color(red)(int" " (5 sin theta)^2/(125 cos^3 theta) " " 5 cos theta " "d theta#

Reduce further:

#int " " (5 sin theta)^2/[(5 cos theta)(5^2 cos^2 theta)] " "(5 cos theta) d theta#

#int " " (5 sin theta)^2/[cancel(5 cos theta)(5^2 cos^2 theta)] " "cancel(5 cos theta) d theta#

#int " " [(25 sin^2 theta)/[25 cos^2 theta)] " "d theta#

#int " " [(cancel 25 sin^2 theta)/[cancel 25 cos^2 theta)] " "d theta#

#int " " (tan^2 theta) " "d theta#

Since #color(red)(tan^2 theta = sec^2 theta - 1#,

#int " " (sec^2 theta - 1) " "d theta#

#int " " sec^2 theta " "d theta - int" " 1 " "d theta#

#rArr tan theta - theta + C# #color(blue)(Exp.2#

Using x-substitution

#tan theta = x/sqrt(25-x^2)# and #theta = sin^(-1)(x/5)#

Write #color(blue)(Exp.2# as

#color(blue)(x/[sqrt(25-x^2)]-sin^(-1)(x/5)+C#