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1) 9 S r^3/(4+r^2)^(1/2) steps: u=r^2 du=2rdr=du/2 9 S u/(4+u)^(1/2) du/2 (9/2) S u/(4+u)^(1/2) du s=4+u and ds=du (9/2) S -4+s/(s)^(1/2) my question is why is s=4+u turn into negative4+s?
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write the statement of question again ,iam not following
\[is \it 9 s=\frac{ r ^{3} }{ \left( 4+r ^{2} \right)^{\frac{ 1 }{ 2}}}\]
S is for integration
\[Let I=\int\limits \frac{ r ^{3} }{\sqrt{4+r ^{2}}}dr\] \[put \sqrt{4+r ^{2}}=u,4+r ^{2}=u ^{2},r ^{2}=u ^{2}-4,2rdr=2udu,rdr=udu\] \[I=\int\limits \frac{ r ^{2}r dr }{\sqrt{4+r ^{2}} }=\int\limits \frac{ \left( u ^{2}-4 \right)u du }{u }=\int\limits \left( u ^{2}-4 \right)du\]
now you can solve.
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