Question

how to do trigonometric substitution?

could you also demonstrate examples :DDD

@apoorvk

Answer 1

how to integrate? differentiate?

Answer 2

integrate lol

Answer 3

or how to make?

Answer 4

i dont think they have trigonometric substitution in differential?

Answer 5

Example: \[\int\frac{\sqrt{1-x^2}}{x^5}\ \text{d}x\]

Answer 6

yes?

Answer 7

substitutions are possible in every sphere. :p

very well.. let me get you a suitable example. Hmm.

Lets take \[\int\sqrt{\frac{x}{{a-x}}}.dx\]

So in here, the prime slogan is "Remove the Radical Sign!"

\[\text{So, I look at the function, and think, hmm, if I can get an }asin^2{\theta}\] that would make it easy, so I get some form of

\[\sqrt{\frac {asin^2\theta}{a - asin^2\theta}}\]

Hmm okay.

But what happens to the 'dx' thingy?

hmm, cool

\[x = asin^2\theta\]
\[\text{So, } dx = 2asin\theta.cos\theta.d\theta\]

Answer 8

Now the whole integral turns out to:
\[2a\int sin\theta cos\theta\sqrt{tan^2\theta}.d \theta\]
\[=2a \int sin\theta cos\theta \frac {sin\theta}{cos\theta}.d\theta\]
\[=\int sin^2\theta.d\theta\]

now you know, \[cos(2\theta) = 1-2sin^2\theta\]
\[or, sin^2\theta = \frac{1- cos2\theta}{2}\]

SO the complete integral turns out to -->

\[\int \frac{1-cos2\theta}{2}.d\theta\]


Which I believe is pretty easily integral using what we call 'loving' integrand formulas :P

Answer 9

Now there are basic fundas, of what I can do when I notice an integral of the following types:

\[\sqrt{a^2 - x^2} ---> put\text{ }in\text{ } x = asin\theta \text{ }or\text{ }acos\theta\]
\[\sqrt{a^2 + x^2} ---> put\text{ }in\text{ } x = atan\theta \text{ }or\text{ } acot\theta\]
\[\sqrt{ x^2- a^2} ---> put\text{ }in\text{ } x = asec\theta \text{ }or\text{ }acosec\theta\]
\[\sqrt{\frac{a-x}{a+x}} --->put \text{ }in\text{ } x= acos2\theta\]

See if you can visualise why we\re doing this. Any guesses? :p

Answer 10

Get this, Mr. @lgbasallote ??

Answer 11

frankly...no :P

Answer 12

i mean..how do you know what to substitute @apoorvk ??

Answer 13

i also don't know where you got \(\sqrt{\tan^2 \theta}\)

Answer 14

i also don't know how it became \(\int \sin ^2 \theta\) and onwards

i don't know a lot i know =))

Answer 15

\[\frac{asin^2x}{a-asin^2x} =\frac{asin^2x}{a(1-sin^2x)}=\frac{sin^2x}{(1-sin^2x)}\]\[=\frac{sin^2x}{(cos^2x)}=tan^2x\]

Answer 16

oh..why didnt i see that o.O