Question

Find the integration of: $\dfrac{{dy}}{{dx}} = x\sqrt {25 - {x^2}} $

Answer 1

Hint: Use integration by substitution to solve this question. Put $25 - {x^2} = t$

Complete Step by Step Solution:
Here, the given equation is$\dfrac{{dy}}{{dx}} = x\sqrt {25 - {x^2}} $
$ \Rightarrow dy = x\sqrt {25 - {x^2}} dx$ . . . . . (1)
Now, the term of$y$is in one side and the term of$x$is in other side, then
Let us consider
$25 - {x^2} = t$ . . . . . (2)
Now, differentiate equation (2) with respect to $x$
Then $ - 2xdx = dt$
$xdx = \dfrac{{dt}}{{ - 2}}$
Now put the value of $xdx$in equation (1)
We get $dy = \sqrt t \dfrac{{dt}}{{ - 2}}$
Re-arranging it, we get $dy = - \dfrac{1}{2}\sqrt t dt$
Now integrate both sides of the equation with respect to $t$
$ \Rightarrow \int {dy = \int { - \dfrac{1}{2}\sqrt t dt} } $
$ \Rightarrow y = \dfrac{{ - 1}}{2}\int {\sqrt t dt} $ (since, constant term can be taken out of the integration)
$ \Rightarrow y = \dfrac{{ - 1}}{2}\int {{{\left( t \right)}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} dt$
$ \Rightarrow y = \dfrac{{ - 1}}{2} \times {t^{\dfrac{3}{2}}} \times \dfrac{2}{3} + C$ $\left( {\because \int {{x^n}dx = {x^{n + 1}} + C} } \right)$
Where, C is the constant of integration.
$ \Rightarrow y = \dfrac{{ - 1}}{3} \times {t^{\dfrac{3}{2}}} + C$
Now put the value of $'t'$ in the given equation.
$ \Rightarrow y = \dfrac{{ - 1}}{3} \times {(25 - {x^2})^{\dfrac{3}{2}}} + C$
So, this is the required solution of the question.

Note: There is a particular type to solve every problem in integration. It is better to first think about a type, the question is best suited for and then apply that type to solve the question. Without thinking about the type first and trying to solve the integration directly may cause problems.