# Evaluate the integral. {eq}\displaystyle \int_4^6 \dfrac 1 {\sqrt {t^2 - 9}}\ dt {/eq}

## Question:

Evaluate the integral.

{eq}\displaystyle \int_4^6 \dfrac 1 {\sqrt {t^2 - 9}}\ dt {/eq}

## Integration:

The area between the curves can be determined by integration. We can simplify the integrals by using trigonometric and inverse trigonometric identities. Finding a function that yields another function on being differentiated is called integration. Using integration, if {eq}\dfrac{d}{dx}\left( F\left( x \right) \right)=f\left( x \right) {/eq}, then {eq}\displaystyle\int{f\left( x \right)dx}=F\left( x \right)+C {/eq}. Some properties of integrals are given below:

• {eq}\displaystyle\int\limits_{a}^{b}{f\left( x \right)dx}=\displaystyle\int\limits_{a}^{b}{f\left( t \right)dt} {/eq}
• {eq}\displaystyle\int\limits_{a}^{b}{f\left( x \right)dx}=-\displaystyle\int\limits_{b}^{a}{f\left( x \right)dx} {/eq}
• {eq}\displaystyle\int\limits_{a}^{b}{f\left( x \right)dx}=\displaystyle\int\limits_{a}^{c}{f\left( x \right)dx}+\displaystyle\int\limits_{c}^{b}{f\left( x \right)dx} {/eq}

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Consider {eq}t=3\sec u {/eq}, so {eq}dt=3\sec u\tan u du {/eq}.

Substitute {eq}4 {/eq} for {eq}t {/eq} into the equation {eq}t=3\sec u {/eq} and...