Questions from Coroneos' 100 integrals (1 Viewer)

Ppen · May 7, 2009

Can some one help me with Q16.

and also Q17

(i'm really struggling with these 100

)

Timothy.Siu · May 7, 2009

Ppen said:
Can some one help me with Q16.

and also Q17

(i'm really struggling with these 100 )

first one, u can do partial fractions
second one, u'd sub in x=atan @

Ppen · May 7, 2009

Timothy.Siu said:
first one, u can do partial fractions
second one, u'd sub in x=atan @

sry i typed it wrong. Q16 is actually:

azureus88 · May 7, 2009

Ppen said:
sry i typed it wrong. Q16 is actually:

sub x=sin@

Ppen · May 7, 2009

Lol, sorry again! I type Q17 wrong (i really suck at this latex). How would you do:

and Q18 which would be similar:

and Q19,

Thanks.

cutemouse · May 7, 2009

azureus88 said:
sub x=sin@

Or x=1/u

Ppen · May 7, 2009

azureus88 said:
sub x=sin@

x=sin @ works perfectly fine for Q16
thx ^^

Dumbledore · May 7, 2009

17, sub x = a*tan@
18, sub x = a*sin@
19, sub x = a*sec@

badquinton304 · May 9, 2009

100 coroneos integration is one of my favourites. Best set of 100 integrals ever.

harrowed2 · May 28, 2009

Having trouble with Q78

Drongoski · May 28, 2009

harrowed2 said:
Having trouble with Q78

Type out the question.

harrowed2 · May 28, 2009

Q78 $\int_{0}^{\frac{\pi }{2}}\frac{dx}{cos^{2}x+2sin^{2}x}$

cutemouse · May 28, 2009

harrowed2 said:
Q78 $\int_{0}^{\frac{\pi }{2}}\frac{dx}{cos^{2}x+2sin^{2}x}$

Divide top and bottom by (cosx)^2 and then let u=tanx

harrowed2 · May 28, 2009

But how do you sub tan pi/2 ?

Drongoski · May 28, 2009

Q78

$\int_{0}^{\frac{\pi }{2}}\frac{1}{cos^{2}x\ +\ 2sin^{2}x}\ dx\ \ =\ \ \int_{0}^{\frac{\pi }{2}}\frac{\frac{1}{cos^{2}x}}{1\ +\ 2tan^{2}x}\ dx\\ \\ \ =\ \ \int_{0}^{\frac{\pi }{2}}\frac{sec^{2}x}{1\ +\ (\sqrt{2}tanx)^{2}}\ dx\ \ =\ \ \frac{1}{\sqrt{2}}\left [ tan^{-1}\sqrt{2}\ tanx \right ]^{\frac{\pi }{2}}_{0}\\ \\ \ =\ \ \frac{1}{\sqrt{2}}\left [ (tan^{-1}(\sqrt{2}\ tan\frac{\pi }{2})\ -\ tan^{-1}\sqrt{2}\ tan0) \right ]\ \ =\ \ \frac{1}{\sqrt{2}}[\frac{\pi }{2}\ -\ 0]\\ \\ \ =\ \ \frac{\pi }{2\sqrt{2}}$

Wonder if this is right ?

harrowed2 · May 28, 2009

That looks great!! Thanks

Pwnage101 · May 28, 2009

where can i get a copy of the 100 integrals?

MC Squidge · May 28, 2009

The coroneos 4u book

cutemouse · May 28, 2009

My teacher did them all and then gave us his solutions =D

Too bad I had already done them before he gave them to us.

harrowed2 · May 29, 2009

Drongoski said:
Q78

$\int_{0}^{\frac{\pi }{2}}\frac{1}{cos^{2}x\ +\ 2sin^{2}x}\ dx\ \ =\ \ \int_{0}^{\frac{\pi }{2}}\frac{\frac{1}{cos^{2}x}}{1\ +\ 2tan^{2}x}\ dx\\ \\ \ =\ \ \int_{0}^{\frac{\pi }{2}}\frac{sec^{2}x}{1\ +\ (\sqrt{2}tanx)^{2}}\ dx\ \ =\ \ \frac{1}{\sqrt{2}}\left [ tan^{-1}\sqrt{2}\ tanx \right ]^{\frac{\pi }{2}}_{0}\\ \\ \ =\ \ \frac{1}{\sqrt{2}}\left [ (tan^{-1}(\sqrt{2}\ tan\frac{\pi }{2})\ -\ tan^{-1}\sqrt{2}\ tan0) \right ]\ \ =\ \ \frac{1}{\sqrt{2}}[\frac{\pi }{2}\ -\ 0]\\ \\ \ =\ \ \frac{\pi }{2\sqrt{2}}$

Wonder if this is right ?

Still a little unsure.
Can inverse tanx equal pi/2 if the range of inverse tan x
is (-pi/2 , pi/2) ?

Questions from Coroneos' 100 integrals (1 Viewer)

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