Question

evaluate integral 1/root 7-6x-z^2​

Answer 1

Solution:

Given Integral:

$\displaystyle \rm = \int \dfrac{1}{ \sqrt{7 - 6x - {x}^{2} } } \: dx$

Can be written as:

$\displaystyle \rm = \int \dfrac{1}{ \sqrt{16 - 9 - 6x - {x}^{2} } } \: dx$

$\displaystyle \rm = \int \dfrac{1}{ \sqrt{16 - (9 + 6x + {x}^{2} )} } \: dx$

$\displaystyle \rm = \int \dfrac{1}{ \sqrt{16 - ( {3}^{2} + 2 \times 3 \times x + {x}^{2} )} } \: dx$

$\displaystyle \rm = \int \dfrac{1}{ \sqrt{16 - {(x + 3)}^{2} } } \: dx$

$\displaystyle \rm = \int \dfrac{1}{ \sqrt{ {4}^{2} - {(x + 3)}^{2} } } \: dx$

Now, let us assume that:

$\rm \longrightarrow u = x + 3$

$\rm \longrightarrow du = dx$

Therefore, the integral changes to:

$\displaystyle \rm = \int \dfrac{du}{ \sqrt{ {4}^{2} - {u}^{2}} }$

We know that:

$\displaystyle \rm \longrightarrow\int \dfrac{du}{ \sqrt{ {a}^{2} - {u}^{2}} } = \sin^{ - 1} \bigg( \dfrac{u}{a} \bigg) + C$

Using this result, we get:

$\displaystyle \rm = \sin^{ - 1} \bigg( \frac{u}{4} \bigg) + C$

Substitute back u = x + 3, we get:

$\displaystyle \rm = \sin^{ - 1} \bigg( \frac{x + 3}{4} \bigg) + C$

Therefore:

$\displaystyle \rm \longrightarrow \int \dfrac{dx}{ \sqrt{7 - 6x - {x}^{2} } } = \sin^{ - 1} \bigg( \frac{x + 3}{4} \bigg) + C$

★ Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$