Thread: Integrations with Trigonometric Substitutions.

1. Integrations with Trigonometric Substitutions.

I have a few variations of some problems that would make the entire homework a lot easier. However, our teacher never covered some of these problems, and I would appreciate some assistance?

a. Integral of 1/(x^2(x^2 - 1)^(1/2)) dx

b. Integral of (x^3)/((x^2 + 4)^(1/2)) dx

c. Integral of ((9 - x^2)^(1/2))/x^2 dx

This is all under Trigonometric substitutions, where substitutions include those sec, cos, and tan.

:S

For a., that extra x squared is miffing me. B. the variable in the numerator prevents a convenient substitution. C. x squared in the denominator is unlike anything we were taught in class.

Help?

2. Hi! niyati, while waiting for someone more competent to help you, here are some tips:

Usually if the integral contains $\displaystyle \sqrt{a^2-x^2}$, where a is constant, you can use the substitution $\displaystyle x=a\sin{\theta}$

When the integral contains $\displaystyle \sqrt{a^2+x^2}$, try $\displaystyle x=a\tan{\theta}$.

When the integral contains $\displaystyle \sqrt{x^2-a^2}$ try $\displaystyle x=a\sec{\theta}$

It's got to do with the common trigonometric identities. Work through a few problems and you'll see it simplifies very nicely. Trust me.

3. Integration by trig substitution.
Sqrt of something.

Umm, the Pythagorean trig identity,
sin^2(A) +cos^2(A) = 1 ----------(i)
Divide both sides by cos^2(A),
tan^2(A) +1 = sec^2(A) ----------(ii)

a. Integral of 1/(x^2(x^2 - 1)^(1/2)) dx

The "sqrt something" is sqrt[x^2 -1].
So we use (ii), or variations of it,
tan^2(A) = sec^2(A) -1 ----------------(iii)
tanA = sqrt[sec^2(A) -1] ------------------------***

Let x = secU -----------***
So,
----x^2 = sec^2(U)
----dx = secUtanU dU
----sqrt[x^2 -1] = sqrt[sec^2(U) -1] = tanU

= INT.[1 / (sec^2(U)*tanU)]*(secUtanU dU)
= INT.[1 /secU]du
= INT.[cosU]dU
= sinU +C

What is sinU?

x = secU
sec = hypotenuse/adjacent side
So, in the reference right triangle of angle u,
--hypotenuse = x
--adjacent side = 1
--opposite side = sqrt[x^2 -1]
So,
sinU = opp/hyp = sqrt[x^2 -1] / x

Therefore,
= {sqrt[x^2 -1] / x} +C ----------------------answer.

4. b. Integral of (x^3)/((x^2 + 4)^(1/2)) dx

The "sqrt something" is sqrt[x^2 +4].
So we use tan^2(A) +1 = sec^2(A).
secA = sqrt[tan^2(A) +1] -----------------------***

sqrt[x^2 +4]
= sqrt[4(x^2)/4 +4/4]
= 2sqrt[(x/2)^2 +1]

Hence, let x/2 = tanU ------------***
So,
---x = 2tanU
---x^3 = 8tan^3(U)
---dx = 2sec^2(U) dU
---sqrt[x^2 +4] = 2sqrt[(x/2)^2 +1] = 2sqrt[tan^2(U) +1] = 2secU

Then,
INT.[(x^3)/(sqrt(x^2 + 4))]dx
= INT.[(8tan^3(U)) / 2secU]*(2sec^2(U) dU)
= (8)INT.[tan^3(U)secU]dU

Umm, secU dU is not the derivative of tanU.
Neither is secUtanU du.
So, cannot integrate yet.

= (8)INT.[tan^2(U)]*secUtanU du
= (8)INT.[sec^2(U) -1]*dsecU
= 8[(1/3)sec^3(U) -secU] +C

What is secU?
x/2 = tanU
So,
--opp side = x
--adj side = 2
--hypotenuse = sqrt[x^2 +4]
And so,
secU = hyp/adj = sqrt[x^2 +4] / 2

Hence,
= 8[(1/3){sqrt[x^2 +4] /2}^3 -sqrt[x^2 +4] /2] +C
= (1/3)[x^2 +4]^(3/2) -4[x^2 +4]^(1/2) +C --------------answer.

5. c. Integral of ((9 - x^2)^(1/2))/x^2 dx

The "sqrt something" is sqrt[9 -x^2]
So we use
sin^2(A) +cos^2(A) = 1
or variation pof it.
cos^2(A) = 1 -sin^2(A) ---------------***

sqrt[9 -x^2]
= sqrt[9/9 -(x^2)/9]
= 3sqrt[1 -(x/3)^2]

So, let x/3 = sinU ---------------***
Then,
--x = 3sinU
--x^2 = 9sin^2(U)
--dx = 3cosU
--sqrt[9 -x^2] = 3sqrt[1 -(x/3)^2] = 3sqrt[1 -sin^2(U)] = 3cosU

Then,
INT.[sqrt(9 - x^2) / x^2] dx
= INT.[3cosU / 9sin^2(U)] (3cosU du)
= INT.[cos^2(U) / sin^2(U)]du
= INT.[cot^2(U)]dU

6. Originally Posted by niyati
a. Integral of 1/(x^2(x^2 - 1)^(1/2)) dx

b. Integral of (x^3)/((x^2 + 4)^(1/2)) dx
Sometimes, we don't need to apply trig. substitutions. Let's see another methods:

$\displaystyle \int\frac1{x^2\sqrt{x^2-1}}\,dx$

The key is defined by the reciprocal substitution $\displaystyle x=\frac1u\implies dx=-\frac1{u^2}\,du,$ after some simple calculations the integral becomes to

$\displaystyle - \int {\frac{u} {{\sqrt {1 - u^2 } }}\,du} = \int {\frac{{\left( {1 - u^2 } \right)'}} {{2\sqrt {1 - u^2 } }}\,du} = \sqrt {1 - u^2 } + k$

Now back substitute and yields

$\displaystyle \int\frac1{x^2\sqrt{x^2-1}}\,dx=\frac{\sqrt{x^2-1}}x+k.$

--

$\displaystyle \int\frac{x^3}{\sqrt{x^2+4}}\,dx$

It suffices to set $\displaystyle u=\sqrt{x^2+4}$ (you'll see why.)