How do you integrate #int x^2/sqrt(x^2-81)dx# using trigonometric substitution?

Question

Ezra Adams · Accepted Answer

#I=x/2sqrt(x^2-81)+81/2ln|x+sqrt(x^2-81)|+C#

Here,

#I=intx^2/sqrt(x^2-81)dx#

Subst. #x=9secu=>dx=9secutanudu, and secu=x/9#

So,

#I=int (81sec^2u)/sqrt(81sec^2u-81)xx9secutanudu#

#=int(81sec^2u)/(cancel9canceltanu)xxcancel9secucanceltanudu#

#=81intsecu*sec^2udu#

#=81intsqrt(1+tan^2u)*sec^2udu#

Let, #tanu=t=>sec^2udu=dt#

#:.I=81intsqrt(1+t^2)dt#

#=81[t/2sqrt(1+t^2)+1^2/2ln|t+sqrt(1+t^2)|]+c#

Subst. back , #t=tanu#

#I#=#81[tanu/2sqrt(1+tan^2u)+1/2ln|tanu+sqrt(1+tan^2u)|]+C'#

=#81[sqrt(sec^2u-1)/2xxsecu+1/2ln|sqrt(sec^2u-1)+secu|]+C'#

Again subst. #secu=x/9#

#I=81[sqrt((x^2)/81-1))/2xxx/9+1/2ln|sqrt((x^2/81)-1)+x/9|]+C'#

#=81[sqrt(x^2-81)/(2xx9)*x/9+1/2ln|sqrt(x^2-81)/9+x/9|]+C'#

#=x/2sqrt(x^2-81)+81/2ln|(x+sqrt(x^2-81))/9|+C'# #=x/2sqrt(x^2-81)+81/2[ln|x+sqrt(x^2-81)|-ln9]+C'#

#=x/2sqrt(x^2-81)+81/2ln|x+sqrt(x^2-81)|-81/2ln9+C'#

#=x/2sqrt(x^2-81)+81/2ln|x+sqrt(x^2-81)|+C#

where #C=C'-81/2ln9#

Scarlett Allison · Answer

# x/2sqrt(x^2-81)+81/2ln|(x+sqrt(x^2-81))|+C.#

Kindly refer to the Second Method which does not use the

Trigo. Subst. to solve the Problem.

Prerequisite :

#intsqrt(x^2-a^2)dx=x/2sqrt(x^2-a^2)-a^2/2ln|(x+sqrt(x^2-a^2))|+c_1#,

&, #intdx/sqrt(x^2-a^2)=ln|(x+sqrt(x^2-a^2))|+c_2#,

Suppose that, #I=intx^2/sqrt(x^2-81)dx#.

#:. I=int{(x^2-81)+81}/sqrt(x^2-81)dx#,

#=int{(x^2-81)/sqrt(x^2-81)+81/sqrt(x^2-81)}dx#,

#=intsqrt(x^2-81)dx+81intdx/sqrt(x^2-81)#,

#=x/2sqrt(x^2-81)-81/2ln|(x+sqrt(x^2-81)|#

#+81ln|(x+sqrt(x^2-81)|#.

#:. I=x/2sqrt(x^2-81)+81/2ln|(x+sqrt(x^2-81))|+C#, as

Respected Maganbhai P. has already derived!.

Enjoy Maths.!

Emily Ammerman · Answer

undefined To integrate $ \int \frac{x^2}{\sqrt{x^2 - 81}} \, dx $ using trigonometric substitution, let $ x = 9\sec(\theta) $. Then, $ dx = 9\sec(\theta)\tan(\theta) \, d\theta $.

Substituting these into the integral, we get:

$$ \int \frac{81\sec^2(\theta)}{\sqrt{81\sec^2(\theta) - 81}} \cdot 9\sec(\theta)\tan(\theta) \, d\theta $$

Simplify this expression to:

$$ \int \frac{81\sec^2(\theta)}{\sqrt{81(\sec^2(\theta) - 1)}} \cdot 9\sec(\theta)\tan(\theta) \, d\theta $$

$$ = \int \frac{81\sec^2(\theta)}{\sqrt{81\tan^2(\theta)}} \cdot 9\sec(\theta)\tan(\theta) \, d\theta $$

$$ = \int \frac{81\sec^2(\theta)}{9\tan(\theta)} \cdot 9\sec(\theta)\tan(\theta) \, d\theta $$

$$ = \int 81\sec^2(\theta) \, d\theta $$

$$ = 81\tan(\theta) + C $$

Now, we need to find $ \tan(\theta) $ in terms of $ x $.

From the substitution $ x = 9\sec(\theta) $, we know that $ \sec(\theta) = \frac{x}{9} $.

Using the identity $ \tan^2(\theta) = \sec^2(\theta) - 1 $, we find $ \tan(\theta) = \sqrt{\sec^2(\theta) - 1} $.

Substituting $ \sec(\theta) = \frac{x}{9} $ into $ \tan(\theta) = \sqrt{\sec^2(\theta) - 1} $, we get:

$$ \tan(\theta) = \sqrt{\left(\frac{x}{9}\right)^2 - 1} $$

$$ \tan(\theta) = \sqrt{\frac{x^2}{81} - 1} $$

$$ \tan(\theta) = \sqrt{\frac{x^2 - 81}{81}} $$

$$ \tan(\theta) = \frac{\sqrt{x^2 - 81}}{9} $$

So, the integral becomes:

$$ \int \frac{x^2}{\sqrt{x^2 - 81}} \, dx = 81\tan(\theta) + C = 81\left(\frac{\sqrt{x^2 - 81}}{9}\right) + C = 9\sqrt{x^2 - 81} + C $$

Therefore, $ \int \frac{x^2}{\sqrt{x^2 - 81}} \, dx = 9\sqrt{x^2 - 81} + C $.