# Complete the square and give a substitution (not necessarily trigonometric) which could be used...

## Question:

Complete the square and give a substitution (not necessarily trigonometric) which could be used to compute the integral:

a. {eq}\int \frac {dy}{y^2+3y+3} {/eq}

b. {eq}\int(2- \theta) cos (\theta^2 - 4 \theta) d \theta {/eq}

## U-Substitutions

The integral given as {eq}\displaystyle \int f(g(x))\cdot g'(x)dx {/eq}

is evaluated using the substitution

{eq}\displaystyle g(x)=u, g'(x)dx=du, {/eq}

therefore the integral is changed to

{eq}\displaystyle \int f(g(x))\cdot g'(x)dx =\int f(u)du. {/eq}

Sometimes, we need to complete the square, by using the following formula {eq}\displaystyle x^2+2ax=(x^2+2ax +a^2)-a^2=(x+a)^2-a^2. {/eq}

## Answer and Explanation:

a. To use the substitution for the integral {eq}\displaystyle \int \frac {dy}{y^2+3y+3} {/eq} we will complete the square for the denominator.

{eq}\displaystyle \begin{align}y^2+3y+3& =y^2+3y+\color{blue}{\left(\frac{3}{2}\right)^2}- \color{blue}{\left(\frac{3}{2}\right)^2}+3 \\ &=\left(y+\frac{3}{2}\right)^2-\frac{9}{4}+3 = \left(y+\frac{3}{2}\right)^2+\frac{3}{4}. \end{align} {/eq}

So, the integral is written as

{eq}\displaystyle \begin{align}\int \frac {dy}{y^2+3y+3}&=\int \frac {dy}{\left(y+\frac{3}{2}\right)^2+\frac{3}{4}}\\ &\implies \boxed{\text{using the substitution } u=y+\frac{3}{2}, du=dy}\implies\\ \int \frac {dy}{y^2+3y+3}&=\int \frac {du}{u^2+\frac{3}{4}}. \end{align} {/eq}

b. For the integral {eq}\displaystyle \int(2- \theta) \cos (\theta^2 - 4 \theta) d \theta {/eq} we don't need to complete the square to obtain a substitution, because outside of the cosine function we have the derivative of the argumetn of the cosine.

So, to solve the integral, we will be using the following substitution {eq}\displaystyle\boxed{ \theta^2 -4\theta =u, (2\theta -4)d\theta=du} \implies (\theta -2)d\theta =\frac{du}{2} {/eq}

and the integral becomes

{eq}\displaystyle \int(2- \theta) \cos (\theta^2 - 4 \theta) d \theta= -\int( \theta-2) \cos (\theta^2 - 4 \theta) d \theta=-\frac{1}{2}\int \cos u\ du. {/eq}

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