Question

Solve: \[\dfrac{{dy}}{{dx}} = {\cos ^3}x{\sin ^2}x + x\sqrt {2x + 1} \]

Answer 1

Hint: We use a variable separable method to separate the variables, and then we take two parts of integrals and solve the parts individually using the substitution method and applying some basic trigonometric properties. After solving the two integrals separately we re-substitute the value of variables and then we add the value of both integrals to get the required solution.

Complete step by step answer:

As the given equation is \[\dfrac{{dy}}{{dx}} = {\cos ^3}x{\sin ^2}x + x\sqrt {2x + 1} \]
As the equation can be separated by variables, so we have
\[ \Rightarrow \]\[dy = \left( {{{\cos }^3}x{{\sin }^2}x + x\sqrt {2x + 1} } \right)dx\]
On integrating both the sides we have,
\[ \Rightarrow \]\[\int {dy} = \int {\left( {{{\cos }^3}x{{\sin }^2}x + x\sqrt {2x + 1} } \right)dx} \]
\[ \Rightarrow \]\[\int {dy} = \int {{{\cos }^3}x{{\sin }^2}x.dx + \int {x\sqrt {2x + 1} .dx} } \]
So, we can have \[y = {I_1} + {I_2}\]
As the given equation can be integrated in the two parts
So, on integrating the first part,
\[ \Rightarrow \]\[{I_1} = \int {{{\cos }^3}x{{\sin }^2}x.dx} \]
Using, \[{\sin ^2}x + {\cos ^2}x = 1\], we have,
\[ \Rightarrow \]\[{I_1} = \int {\cos x{{\sin }^2}x\left( {1 - {{\sin }^2}x} \right).dx} \]
Now let,
\[
  \sin x = t \\
  \cos x.dx = dt \\
  \]
Hence, substituting the above terms, we get,
\[ \Rightarrow \]\[{I_1} = \int {{t^2} - {t^4}.dt} \]
On integrating, we get,
\[ \Rightarrow \]\[{I_1} = \dfrac{{{t^3}}}{3} - \dfrac{{{t^5}}}{5} + c\]
Again, substituting the value of t, we get,
\[ \Rightarrow \]\[{I_1} = \dfrac{{{{\sin }^3}x}}{3} - \dfrac{{{{\sin }^5}x}}{5} + c\]…………..(1)
On integrating the second part of integration as,
\[{I_2} = \int {{{\left( {2x + 1} \right)}^{\dfrac{1}{2}}}.x.dx} \]
Take \[2x + 1 = {v^2}\]
Take differentiation on both sides as \[2.dx = 2v.dv\],
On simplification we get,
\[dx = vdv\]
Also, on simplification of \[2x + 1 = {v^2}\], we get,
\[x = \dfrac{{{v^2} - 1}}{2}\]
On putting all the values in above integral we get,
\[ \Rightarrow \]\[{I_2} = \int {\dfrac{{{v^2} - 1}}{2}\left( v \right)v.dv} \]
On simplification,
\[ \Rightarrow \]\[{I_2} = \dfrac{1}{2}\int {{v^4} - {v^2}.dv} \]
On integrating, we get,
\[ \Rightarrow \]\[{I_2} = \dfrac{1}{2}\left( {\dfrac{{{v^5}}}{5} - \dfrac{{{v^3}}}{3}} \right) + c\]
On simplifying, we get,
\[ \Rightarrow \]\[{I_2} = \dfrac{{{v^5}}}{{10}} - \dfrac{{{v^3}}}{6} + c\]
Put the value v again back from to second part of integral as,
\[{I_2} = \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{5}{2}}}}}{{10}} - \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{3}{2}}}}}{6} + c\]
Hence, on adding both the integral, we get,
\[y = \dfrac{{{{\sin }^3}x}}{3} - \dfrac{{{{\sin }^5}x}}{5} + \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{5}{2}}}}}{{10}} - \dfrac{{{{\left( {2x + 1} \right)}^{\dfrac{3}{2}}}}}{6} + c\]
Hence, above is our required answer.

Note: Use the various integration concept such as of substitution, normal integration as using all that in the above integral and
Using \[dy = \int {f.dx} \]as general concept. Hence, the answer required can be obtained.
In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. Integration is one of the two main operations of calculus; its inverse operation, differentiation, is the other.