Use an appropriate substitution and then trigonometric substitution to evaluate the following...

Question:

Use an appropriate substitution and then trigonometric substitution to evaluate the following integral:

{eq}I = \int \frac {x}{\sqrt {x^2 + x + 1)}} {/eq}

Integration By Substitution.

U-substitution method gives a simpler integration with involving a variable u.

It is a way of representing the function with respect to another function.

If we want to change the limit of the integration we can apply the substitution method.

The formula is:

{eq}\displaystyle\int secx\ dx=ln\left | tanx+secx \right |+c\\\\ \displaystyle\int x^n\ dx=\dfrac{x^{n+1}}{n+1}+c\\\\ {/eq}

Answer and Explanation:

Here we have to use substitution method to evaluate the integrals:

{eq}I=\displaystyle\int \dfrac{x}{\sqrt {x^2+x+1}}dx\\\\ {/eq}

We may write as:

{eq}x^2+x+1=(x+\dfrac{1}{2})^2+\dfrac{3}{4}\\\\ {/eq}

Hence:

{eq}\displaystyle\int \dfrac{x}{\sqrt {\left ( x+\dfrac{1}{2} \right )^2+\dfrac{3}{4}}}dx\\\\ {/eq}

Let:

{eq}x+\dfrac{1}{2}=z\\\\ dx=dz\\\\ {/eq}

Or,

{eq}x=z-\dfrac{1}{2}\\\\ {/eq}

Therefore:

{eq}\displaystyle\int \dfrac{2z-1}{2\sqrt {z^2+\dfrac{3}{4}}}dz\\\\ \displaystyle\int \dfrac{2z-1}{\sqrt {4z^2+3}}dz\\\\ \displaystyle\int \dfrac{2z}{\sqrt {4z^2+3}}dz-\displaystyle\int \dfrac{1}{\sqrt {4z^2+3}}dz\,\,\,\,\,.............eqn(1)\\\\ {/eq}

Now integrating separately the first part:

{eq}\displaystyle\int \dfrac{2z}{\sqrt {4z^2+3}}dz\\\\ {/eq}

Let:

{eq}4z^2+3=t\\\\ 4\cdot 2z\ dz=dt\\\\ 2z\ dz=\dfrac{dt}{4}\\\\ {/eq}

Hence:

{eq}\displaystyle\int \dfrac{1}{\sqrt t}\cdot \dfrac{dt}{4}\\\\ =\dfrac{1}{4}\left [ \dfrac{t^{\frac{1}{2}}}{\frac{1}{2}} \right ]\\\\ =\dfrac{2}{4}\left [ \sqrt {4z^2+3} \right ]\\\\ =\dfrac{1}{2}\sqrt {4z^2+3}\\\\ {/eq}

And also integrating the second part:

{eq}\displaystyle\int \dfrac{1}{\sqrt {4z^2+3}}dz\\\\ {/eq}

Let:

{eq}z=\dfrac{\sqrt 3}{2}tanv\\\\ dz=\dfrac{\sqrt 3}{2}sec^2v\ dv\\\\ {/eq}

Or,

{eq}v=tan^{-1}\left ( \dfrac{2z}{\sqrt 3} \right )\\\\ {/eq}

Therefore:

{eq}\displaystyle\int \dfrac{1}{\sqrt {4\cdot \dfrac{3}{4}tan^2v+3}}\cdot \dfrac{\sqrt 3}{2}sec^2v\ dv\\\\ =\displaystyle\int \dfrac{1}{\sqrt 3\sqrt {tan^2v+1}}\cdot \dfrac{\sqrt 3}{2}sec^2v\ dv\\\\ =\dfrac{1}{2}\displaystyle\int \dfrac{1}{secv}\cdot sec^2v\ dv\\\\ =\dfrac{1}{2}\displaystyle\int secv\ dv\\\\ =\dfrac{1}{2}ln\left | tan(v)+sec(v) \right |+c\\\\ =\dfrac{1}{2}ln\left | tan\left ( tan^{-1}\left ( \dfrac{2z}{\sqrt 3} \right ) \right )+sec\left ( tan^{-1}\left ( \dfrac{2z}{\sqrt 3} \right ) \right ) \right |+c\\\\ =\dfrac{1}{2}ln\left [ \dfrac{2}{\sqrt 3}\left ( x+\dfrac{1}{2} \right )+sec\left [ tan^{-1}\left ( \dfrac{2}{\sqrt 3}\left ( x+\dfrac{1}{2} \right ) \right ) \right ] \right ]+c\\\\ {/eq}

Substituting the above values in eqn(1) we get:

{eq}=\dfrac{1}{2}\sqrt {4\left ( x+\dfrac{1}{2} \right )^2+3}-\left [ \dfrac{1}{2}ln\left [ \dfrac{2}{\sqrt 3}\left ( x+\dfrac{1}{2} \right )+sec\left [ tan^{-1}\left ( \dfrac{2}{\sqrt 3}\left ( x+\dfrac{1}{2} \right ) \right ) \right ] \right ] \right ]+c\\\\ {/eq}

Here c is integration constant.


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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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