How do you integrate #int x^3/sqrt(x^2+10)dx# using trigonometric substitution?

Answer 1

#int x^3/sqrt(x^2+10)dx = (sqrt(x^2+10)(x^2-20))/3 +C#

Substitute:

#x=sqrt10tant#
#dx = sqrt10sec^2tdt#
with #t in (-pi/2,pi/2)#.

So:

#int x^3/sqrt(x^2+10)dx = int (10sqrt10tan^3tsqrt10sec^2t)/sqrt(10tan^2t+10)dt#
#int x^3/sqrt(x^2+10)dx = int (100tan^3tsec^2t)/(sqrt10 sqrt(tan^2t+1))dt#
#int x^3/sqrt(x^2+10)dx = 10sqrt10 int (tan^3tsec^2t)/ sqrt(tan^2t+1)dt#

Use now the trigonometric identity:

#tan^2t +1 = sec^2t #
note that for #t in (-pi/2,pi/2)# the secant is positive, so:
#sqrt(tan^2t +1) = sect #

and:

#int x^3/sqrt(x^2+10)dx = 10sqrt10 int (tan^3tsec^2t)/ sectdt#
#int x^3/sqrt(x^2+10)dx = 10sqrt10 int tan^3tsect dt#
Expand now in terms of #sint# and #cost#:
#int x^3/sqrt(x^2+10)dx = 10sqrt10 int sin^3t /cos^3t 1/cost dt#
#int x^3/sqrt(x^2+10)dx = 10sqrt10 int (1-cos^2t) /cos^4t sint dt#

Substitute now:

#u = cost#
#du = -sint dt#

to have:

#int x^3/sqrt(x^2+10)dx = -10sqrt10 int (1-u^2) /u^4 du#

and using the linearity of the integral:

#int x^3/sqrt(x^2+10)dx = -10sqrt10( int (du)/u^4 + int (du) /u^2) #
#int x^3/sqrt(x^2+10)dx = 10sqrt10( 1/(3u^3) -1/u) +C#

undoing the substitution:

#int x^3/sqrt(x^2+10)dx = 10/3 1/cos^3t -10/cost +C#
#int x^3/sqrt(x^2+10)dx = (10sqrt10)/3(sec^3t -3sect) +C#
#int x^3/sqrt(x^2+10)dx = (10sqrt10sect)/3(sec^2t -3) +C#
#int x^3/sqrt(x^2+10)dx = 10sqrt10sect((sec^2t -1) -2) +C#
#int x^3/sqrt(x^2+10)dx = 10sqrt10 sqrt(1+tan^2t)(tan^2t -2) +C#
#int x^3/sqrt(x^2+10)dx = (10sqrt10)/3sqrt(1+x^2/10)(x^2/10-2) +C#
#int x^3/sqrt(x^2+10)dx = (sqrt(x^2+10)(x^2-20))/3 +C#

I would note however that even if the question required integration by trigonometric substitution, it would be easier to integrate differently:

#int x^3/sqrt(x^2+10)dx = int x^2 (xdx)/sqrt(x^2+10)#

Substitute:

#u = sqrt(x^2+10)#
#du = (xdx)/sqrt(x^2+10)#
#x^2 = u^2-10#

so:

#int x^3/sqrt(x^2+10)dx = int (u^2-10) du#
#int x^3/sqrt(x^2+10)dx = int u^2du -10int du#
#int x^3/sqrt(x^2+10)dx = u^3/3 -10u+C#
#int x^3/sqrt(x^2+10)dx = u/3 (u^2-30)+C#
#int x^3/sqrt(x^2+10)dx = (sqrt(x^2+10)(x^2-20))/3+C#
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Answer 2

To integrate ( \int \frac{x^3}{\sqrt{x^2+10}} , dx ) using trigonometric substitution, let ( x = \sqrt{10} \tan(\theta) ). Then, ( dx = \sqrt{10} \sec^2(\theta) , d\theta ). Substituting these into the integral gives:

[ \int \frac{x^3}{\sqrt{x^2+10}} , dx = \int \frac{(\sqrt{10} \tan(\theta))^3}{\sqrt{(\sqrt{10} \tan(\theta))^2+10}} (\sqrt{10} \sec^2(\theta)) , d\theta ]

[ = \int \frac{10\sqrt{10}\tan^3(\theta)}{\sqrt{10\tan^2(\theta)+10}} \cdot \sqrt{10}\sec^2(\theta) , d\theta ]

[ = \int \frac{10\sqrt{10}\tan^3(\theta)}{\sqrt{10(\tan^2(\theta)+1)}} \cdot \sqrt{10}\sec^2(\theta) , d\theta ]

[ = \int \frac{10\sqrt{10}\tan^3(\theta)}{\sqrt{10}\sec(\theta)} \cdot \sqrt{10}\sec^2(\theta) , d\theta ]

[ = \int 10\tan^3(\theta) \cdot \sec(\theta) , d\theta ]

Now, we can use the trigonometric identity ( \tan^2(\theta) = \sec^2(\theta) - 1 ) to simplify the integral:

[ = \int 10(\sec^2(\theta) - 1)\tan(\theta) \cdot \sec(\theta) , d\theta ]

[ = \int 10(\sec^3(\theta) - \sec(\theta)) , d\theta ]

[ = 10\int \sec^3(\theta) , d\theta - 10\int \sec(\theta) , d\theta ]

The integral ( \int \sec^3(\theta) , d\theta ) can be evaluated using the reduction formula for integrals of powers of secant. The integral ( \int \sec(\theta) , d\theta ) is a standard integral.

After finding these integrals and substituting back ( x = \sqrt{10} \tan(\theta) ), you will have the result in terms of ( x ) and ( \theta ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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