How do you integrate int -x^2/sqrt(144+x^2)dx using trigonometric substitution?

Jul 16, 2018

$\implies I = 72 \ln | x + \sqrt{{x}^{2} + 144} | - \frac{x}{2} \sqrt{{x}^{2} + 144} + C$

Explanation:

Here ,

$I = - \int {x}^{2} / \sqrt{{x}^{2} + 144} \mathrm{dx}$

Subst. color(blue)(x=12tanu=>dx=12sec^2udu and

color(blue)(tanu=x/12=>secu=sqrt(1+tan^2u)=sqrt(1+x^2/144)

=>color(blue)(secu=sqrt(144+x^2)/12

So,

$I = - \int \frac{144 {\tan}^{2} u}{\sqrt{144 {\tan}^{2} u + 144}} \cdot 12 {\sec}^{2} u \mathrm{du}$

$\implies I = - 144 \int \frac{{\tan}^{2} u}{12 \sec u} 12 {\sec}^{2} u \mathrm{du}$

$\implies I = - 144 \int \sec u {\tan}^{2} u \mathrm{du}$

$\implies I = - 144 \cdot {I}_{1} - - - - \to \left(A\right)$

$W h e r e , {I}_{1} = \int \sec u {\tan}^{2} u \mathrm{du} \ldots \to \left(B\right)$

=>I_1=intsecu(sec^2u-1)du...to[becausesec^2theta- tan^2theta=1]

$\implies {I}_{1} = \int \sec u {\sec}^{2} u \mathrm{du} - \int \sec u \mathrm{du}$

Using Integration by Parts in first integral

${I}_{1}$=$\left\{\sec u \int {\sec}^{2} u - \int \left(\sec u \tan u \int {\sec}^{2} u \mathrm{du}\right) \mathrm{du}\right\} - \int \sec u \mathrm{du}$

${I}_{1}$=$\left\{\sec u \cdot \tan u - \int \sec u \tan u \cdot \tan u \mathrm{du}\right\} - \ln | \sec u + \tan u |$

$\implies {I}_{1} = \sec u \tan u - \int \sec u {\tan}^{2} u \mathrm{du} - \ln | \sec u + \tan u |$

$\implies {I}_{1} = \sec u \tan u - {I}_{1} - \ln | \sec u + \tan u | + c \ldots \to \left[u s e , \left(B\right)\right]$

$\implies 2 {I}_{1} = \sec u \tan u - \ln | \sec u + \tan u | + c$
Subst. back , color(blue)(tanu=x/12 and secu=sqrt(144+x^2)/12

$2 {I}_{1} = \frac{\sqrt{144 + {x}^{2}}}{12} \cdot \frac{x}{12} - \ln | \frac{\sqrt{144 + {x}^{2}}}{12} + \frac{x}{12} | + c$

$\implies 2 {I}_{1} = \frac{x}{144} \sqrt{144 + {x}^{2}} - \left\{\ln | \sqrt{144 + {x}^{2}} + x | - \ln 12\right\} + c$

$\implies {I}_{1} = \frac{1}{2} \cdot \frac{x}{144} \sqrt{144 + {x}^{2}} - \frac{1}{2} \ln | \sqrt{144 + {x}^{2}} + x | + C '$

$w h e r e , C ' = \frac{1}{2} \left(c - \ln 12\right)$

Now ,from $\left(A\right)$ we get

$I$=$- 144 \left\{\frac{1}{2} \cdot \frac{x}{144} \sqrt{144 + {x}^{2}} - \frac{1}{2} \ln | x + \sqrt{144 + {x}^{2}} |\right\} + C$

$I = - \frac{1}{2} x \sqrt{144 + {x}^{2}} + 72 \ln | x + \sqrt{{x}^{2} + 144} | + C$

$\implies I = 72 \ln | x + \sqrt{{x}^{2} + 144} | - \frac{x}{2} \sqrt{{x}^{2} + 144} + C$

Jul 16, 2018

The answer is $= - 6 x \sqrt{1 + {x}^{2} / 144} + 72 \ln \left(| \frac{x}{12} + \sqrt{1 + {x}^{2} / 144} |\right) + C$

Explanation:

The integral is

$I = \int - \frac{{x}^{2} \mathrm{dx}}{\sqrt{144 + {x}^{2}}}$

Rewrite it as

$I = \int - \frac{\left({x}^{2} + 144 - 144\right) \mathrm{dx}}{\sqrt{144 + {x}^{2}}}$

$= \int \frac{144 \mathrm{dx}}{\sqrt{144 + {x}^{2}}} - \int \frac{\left(144 + {x}^{2}\right) \mathrm{dx}}{\sqrt{144 + {x}^{2}}}$

$= \int \frac{144 \mathrm{dx}}{\sqrt{144 + {x}^{2}}} - \int \sqrt{144 + {x}^{2}} \mathrm{dx}$

$= {I}_{1} + {I}_{2}$

Compute the $2 n d$ integral by substitution

Let $x = 12 \tan \theta$

$\mathrm{dx} = 12 {\sec}^{2} \theta d \theta$

$\sqrt{144 + {x}^{2}} = \sqrt{144 + 144 {\tan}^{2} \theta} = 12 \sec \theta$

Therefore,

${I}_{2} = \int \sqrt{144 + {x}^{2}} \mathrm{dx} = 144 \int {\sec}^{3} \theta d \theta = 144 \int {\sec}^{2} \theta \sec \theta d \theta$

Perform this integration by parts

$u = \sec \theta$, $\implies$, $u ' = \sec \theta \tan \theta$

$v ' = {\sec}^{2} \theta$, $\implies$, $v = \tan \theta$

Therefore,

$144 \int {\sec}^{3} \theta d \theta = 144 \sec \theta \tan \theta - 144 \int \sec \theta {\tan}^{2} \theta d \theta$

$= 144 \sec \theta \tan \theta - 144 \int \left({\sec}^{2} \theta - 1\right) \sec \theta d \theta$

$= 144 \sec \theta \tan \theta - 144 \int {\sec}^{3} \theta d \theta + 144 \int \sec \theta d \theta$

$288 \int {\sec}^{3} \theta d \theta = 144 \sec \theta \tan \theta + 144 \int \sec \theta d \theta$

$= 144 \sec \theta \tan \theta + 144 \ln \left(| \sec \theta + \tan \theta |\right)$

$144 \int {\sec}^{3} \theta d \theta = 72 \sec \theta \tan \theta + 72 \ln \left(| \sec \theta + \tan \theta |\right)$

${I}_{2} = 72 \cdot \frac{x}{12} \sqrt{1 + {x}^{2} / 144} + 72 \ln \left(| \frac{x}{12} + \sqrt{1 + {x}^{2} / 144} |\right)$

Compute the first integral

${I}_{1} = \int \frac{144 \mathrm{dx}}{\sqrt{144 + {x}^{2}}} = 144 \int \frac{\mathrm{dx}}{\sqrt{144 + {x}^{2}}}$

Let $u = \frac{x}{12}$, $\implies$, $\mathrm{du} = \frac{1}{12} \mathrm{dx}$

I_1=144int(12du)/(12sqrt(1+u^2)

$= 144 \int \frac{\mathrm{du}}{\sqrt{1 + {u}^{2}}}$

Let $u = \tan v$, $\implies$, $\mathrm{du} = {\sec}^{2} v \mathrm{dv}$

$\sqrt{1 + {u}^{2}} = \sqrt{1 + {\tan}^{2} v} = \sec v$

${I}_{1} = 144 \int \frac{{\sec}^{2} v \mathrm{dv}}{\sec v} = 144 \int \sec v \mathrm{dv}$

$= 144 \ln \left(| \sec v + \tan v |\right)$

$= 144 \ln \left(\sqrt{1 + {u}^{2}} + u\right)$

$= 144 \ln \left(\sqrt{1 + {x}^{2} / 144} + \frac{x}{12}\right)$

Putting it alltogether

$I = - 6 x \sqrt{1 + {x}^{2} / 144} - 72 \ln \left(\frac{x}{12} + \sqrt{1 + {x}^{2} / 144}\right) + 144 \ln \left(\sqrt{1 + {x}^{2} / 144} + \frac{x}{12}\right)$

$= - 6 x \sqrt{1 + {x}^{2} / 144} + 72 \ln \left(| \frac{x}{12} + \sqrt{1 + {x}^{2} / 144} |\right) + C$

Jul 16, 2018

$- \frac{1}{2} x \sqrt{{x}^{2} + 144} + 72 \ln | \frac{\sqrt{{x}^{2} + 144} + x}{12} | + C$.

Explanation:

Let, $I = \int - \left({x}^{2} / \left(\sqrt{{x}^{2} + 144}\right)\right) \mathrm{dx} = - \int {x}^{2} / \sqrt{{x}^{2} + 144} \mathrm{dx}$.

We subst. $x = 12 \tan y . \text{ Then, } \mathrm{dx} = 12 {\sec}^{2} y \mathrm{dy}$.

$\therefore I = - \int \frac{\left(144 {\tan}^{2} y\right) \left(12 {\sec}^{2} y\right)}{\sqrt{144 {\tan}^{2} y + 144}} \mathrm{dy}$,

$= - 144 \int \sec y {\tan}^{2} y \mathrm{dy}$,

$= - 144 \int \left(\tan y\right) \left(\sec y \tan y \mathrm{dy}\right)$,

$= - 144 \int \sqrt{{\sec}^{2} y - 1} \left(\sec y \tan y \mathrm{dy}\right)$.

So, letting, secy=t, &," so, "secytanydy=dt, we have,

$I = - 144 \int \sqrt{{t}^{2} - 1} \mathrm{dt}$,

$= - 144 \cdot \frac{1}{2} \left\{t \sqrt{{t}^{2} - 1} - \ln | t + \sqrt{{t}^{2} - 1} |\right\}$,

=-72{secytany-ln|secy+tany||,

$= - 72 \left\{\tan y \sqrt{{\tan}^{2} y + 1} - \ln | \sqrt{{\tan}^{2} y + 1} + \tan y |\right\}$,

$= - 72 \left\{\frac{x}{12} \cdot \sqrt{{x}^{2} / 144 + 1} - \ln | \sqrt{{x}^{2} / 144 + 1} + \frac{x}{12} |\right\}$.

$= - 72 \left\{\frac{x \sqrt{{x}^{2} + 144}}{144} - \ln | \frac{\sqrt{{x}^{2} + 144} + x}{12} |\right\}$.

$\Rightarrow I = - \frac{1}{2} x \sqrt{{x}^{2} + 144} + 72 \ln | \frac{\sqrt{{x}^{2} + 144} + x}{12} | + C$.

N.B.:- The above Soln. has been worked out using trigo.**

substn. as it was so expected. But the same can be dealt

with without using any substn. as shown in the Second Soln.

What I had in my mind as the second soln., Respected Narad T.

has solved it exactly in the same way, so, I think, there is

no need for my second soln.

However, I prefer to use the following Standard Integrals :

$\int \frac{1}{\sqrt{{x}^{2} + {a}^{2}}} \mathrm{dx} = \ln | \left(x + \sqrt{{x}^{2} + {a}^{2}}\right) | + {c}_{1} , \mathmr{and} ,$

$\int \sqrt{{x}^{2} + {a}^{2}} \mathrm{dx} = \frac{1}{2} \left\{x \sqrt{{x}^{2} + {a}^{2}} + {a}^{2} \ln | \left(x + \sqrt{{x}^{2} + {a}^{2}}\right) |\right\} + {c}_{2}$.