Inverse trig integration (1 Viewer)

[kaz1]

 

[lychnobity]

let x = 3sina, dx = 3cosa da
So this: turns into this:

=∫3cosa.3cosa da
=∫9cos²a da
=9/2∫1 + cos2a da
= 9a/2 + 9sin2a/4+ c

my bad, fixed.

 

[Dumbledore]

let x = 3sina, dx = 3cosa da
So this: turns into this:

=∫3cosa.3cosa da
=∫9cos²a da
=9/2∫1 - cos2a da
= 9x/2 - 9sin2a/4+ c

i think cos^2(x) = (1+cos2x)/2

 

[kwabon]

why did you label the thread's name as inverse trig??? because the integration used for this is just normal trig. if you want us to use inverse trig for this type of integral, lol, i have no clue.

 

[clintmyster]

i think cos^2(x) = (1+cos2x)/2

yep your right

 

[kaz1]

How do you know when to make a trigonometric substitution?

 

[lyounamu]

How do you know when to make a trigonometric substitution?

some questions just appear that way and if you have a question like that you usually are expected to make that sub anyway.

 

[addikaye03]

i got 1/2[sqrt(9-x^2)+9sin^-1(x/3)]+C

 

[Trebla]

How do you know when to make a trigonometric substitution?

Because no other substitution works as nicely. Alternatively, you could try integration by parts which gives a recurrence of integral if you know how to manipulate it properly.

 

[Aquawhite]

You will often notice that almost all trig substituions have either different of two squares or something with perfect squares in them to make them a lot easier (thus making them a bit easier to identify...)

 

[Dumbledore]

i got 1/2[sqrt(9-x^2)+9sin^-1(x/3)]+C

yeh thats probably the same, previous answer just didn't substitute back to get in terms of x

 

[clintmyster]

Because no other substitution works as nicely. Alternatively, you could try integration by parts which gives a recurrence of integral if you know how to manipulate it properly.

my teacher told us something similar

as a last resort you always use either trig sub, t sub or integration by parts.

 

[addikaye03]

yeh thats probably the same, previous answer just didn't substitute back to get in terms of x

Yer i didn't see the above attempt. That was what i did, Just done the next step (solved for x)

 

[Aerath]

How do you know when to make a trigonometric substitution?

Generally, when you see f(a^2-x^2), use substitution x = asin@ (or acos@)
f(a^2 + x^2) use x = atan@
f(x^2-a^2) use x = asec@

I generally try to avoid using substitutions...cause it can get very messy, so watch out for the quarter or semicircles, cause they appear often.

 

[random-1005]

How do you know when to make a trigonometric substitution?

when you have a difference of two squares under a squareroot

 

[clintmyster]

Generally, when you see f(a^2-x^2), use substitution x = asin@ (or acos@)
f(a^2 + x^2) use x = atan@
f(x^2-a^2) use x = asec@

I generally try to avoid using substitutions...cause it can get very messy, so watch out for the quarter or semicircles, cause they appear often.

provided there are limits ofcourse. This method however is very useful for avoiding the necessary trig sub in volumes questions.

 

[tehrobzorz]

when you have a difference of two squares under a squareroot

also simply when there is a fraction involving squares

like 1/(x^2+a^2)^3 dx

or when you need to rationalise

like sqrt[(5-x)/(x-1)] dx

but usually it involves a square root. BUT! becareful cause ive come across questions where you think you need to complete the square and then trig sub.. but it ends up simply ending up as f'(x)/f(x) = ln(x) so yeah =X