# Find the indefinite integral \int x^5 \sin^2 x(x \cos x + 2 \sin x) \sqrt{x^4 \sin^2 x + 5} dx...

## Question:

Find the indefinite integral {eq}\displaystyle \int x^5 \sin^2 x(x \cos x + 2 \sin x) \sqrt{x^4 \sin^2 x + 5} \ dx {/eq} using the integral table. Cite the formula(e) and substitution used.

## Integration:

Integration is a mathematical tool that helps us to calculate the area of the curve, central points, volume, and other real-world applications. When the area of a function is calculated by combining small elemental slices, then it is called integration.

Given integral: {eq}\int {{x^5}{{\sin }^2}(x)(x \cdot cos(x) + 2 \cdot sin(x))\sqrt {{x^4}{{\sin }^2}(x) + 5} } \;dx{/eq}

{eq}\begin{align*} {\rm{substitute }}\;t &= {x^2}sin(x) \to dt = 2x\sin (x) + {x^2}\cos (x)dx\\ {\rm{using }}\;{t^2} &= {x^4}si{n^2}(x),\;\\{\rm{Our \;integral \;becomes:\;}}\;\\ &= \int {{t^2}\sqrt {{t^2} + 5} \;dt} \end{align*} {/eq}

Now we will perform hyperbolic substitution and our integral becomes:

{eq}{\rm{substitute}}\;t = \sqrt 5 \sinh (u) \to dt = \sqrt 5 \cosh (u)du{/eq}

{eq}\begin{align*} = \int {{5^{\dfrac{3}{2}}}} \cosh (u)sin{h^2}(u)\sqrt {5{{\sinh }^2}(u) + 5} \;du\\ \end{align*} {/eq}

{eq}\begin{align*} {\rm{simplify \;using:}}\;&{\rm{5sin}}{{\rm{h}}^2}(u) + 5 = 5({\rm{sin}}{{\rm{h}}^2}(u) + 1) = 5cos{h^2}(u) \end{align*} {/eq}

Integral becomes:

{eq}= 25 \cdot \int {{{\cosh }^2}(u)sin{h^2}(u)du} {/eq}

Now we will use product-sum formulas for the generated hyperbolic function to solve the integral:

{eq}\begin{align*} {\sinh ^2}(x) &= \dfrac{1}{2}\left( {\cosh (2x) - 1} \right)\\ {\cosh ^2}(x) &= \dfrac{1}{2}\left( {\cosh (2x) - 1} \right) \end{align*} {/eq}

On simplifying, we get;

{eq}\begin{align*} &= \dfrac{1}{2}\left( {\cosh (2u) - 1} \right) \cdot \dfrac{1}{2}\left( {\cosh (2u) - 1} \right)\\ &= \dfrac{1}{4}({\cosh ^2}(2u) - 1)\\ &= \dfrac{1}{4}\left( {\dfrac{1}{2}\left( {\cosh (4u) - 1} \right) - 1} \right)\\ &= \dfrac{1}{8}\left( {\cosh (4u) - 1} \right) \end{align*} {/eq}

So, integration now becomes:

{eq}\begin{align*} &= \dfrac{1}{8}\int {\cosh (4u) - 1} \;du\\ &= \dfrac{1}{8}\int {\cosh (4u)\;du - } \dfrac{1}{8}\int {1\;du} \\ \end{align*} {/eq}

{eq}\begin{align*} {\rm{Substitute \;}}v = 4u \to du = \dfrac{1}{4}dv \end{align*} {/eq}

Now the integral changes to:

{eq}\begin{align*} &= \dfrac{1}{8}\int {\dfrac{1}{4}\cosh (v)\;dv} - \dfrac{1}{8}\int {1\;du} \\ &= \dfrac{1}{8}\left( {\dfrac{1}{4}[sinh(v)]} \right) - \dfrac{1}{8}[u]\\ &= \dfrac{1}{8}\left( {\dfrac{{sinh(4u)}}{4}} \right) - \dfrac{1}{8}[u]\\ &= \dfrac{{sinh(4u)}}{{32}} - \dfrac{u}{8} \end{align*} {/eq}

Then we will plug back the value and undo the substitution {eq}u = sin{h^{ - 1}}\left( {\dfrac{t}{{\sqrt 5 }}} \right) {/eq}and then for {eq}t = {x^2}\sin (x){/eq} in the integral:

{eq}\begin{align*} &= 25\left( {\dfrac{{sinh\left( {4sin{h^{ - 1}}\left( {\dfrac{{{x^2}\sin (x)}}{{\sqrt 5 }}} \right)} \right)}}{{32}} - \dfrac{{sin{h^{ - 1}}\left( {\dfrac{{{x^2}\sin (x)}}{{\sqrt 5 }}} \right)}}{8}} \right)\\ &= \dfrac{{25sinh\left( {4sin{h^{ - 1}}\left( {\dfrac{{{x^2}\sin (x)}}{{\sqrt 5 }}} \right)} \right)}}{{32}} - \dfrac{{25sin{h^{ - 1}}\left( {\dfrac{{{x^2}\sin (x)}}{{\sqrt 5 }}} \right)}}{8}\\ \end{align*} {/eq}

After simplification and substituting back variables, the indefinite integration can be given as:

{eq}\int {{x^5}{{\sin }^2}(x)(x \cdot cos(x) + 2 \cdot sin(x))\sqrt {{x^4}{{\sin }^2}(x) + 5} } \;dx = \dfrac{{25sinh\left( {4sin{h^{ - 1}}\left( {\dfrac{{{x^2}\sin (x)}}{{\sqrt 5 }}} \right)} \right)}}{{32}} - \dfrac{{25sin{h^{ - 1}}\left( {\dfrac{{{x^2}\sin (x)}}{{\sqrt 5 }}} \right)}}{8} + C {/eq}