Question

INTEGRATION INVOLVING TRIG

Answer 1

\[\int \sqrt{1+sin^2(x-1)}sin(x-1)cos(x-1)dx\] is the question ....

Answer 2

Let u = 1+sin^2 (x-1) => du = 2sin(x-1)cos(x-1) dx
\[\int \sqrt{1+sin^2(x-1)}sin(x-1)cos(x-1)dx \]\[=1/2\int \sqrt{u}du = 1/2 x(2/3) u^{3/2} +C\]= (1/3) [1+sin(x-1)^(3/2) ]+c

Answer 3

Okay in this, just open up the part "sin(x-1)cos(x-1)" and simplify it into the roots.

you 'll get something like

\[{cos2} \int \sqrt{1+sin^2x}sinxcosx.dx - \frac {sin2}{2}\int \sqrt{1+sin^2x}.sin^2x.dx \] \[+\frac {sin2}{2}\int \sqrt{1+sin^2x}(cos^2x).dx\]


And this I believe is integrable quite easily.

Answer 4

Sorry, the last step should be
= (1/3) [1+sin(x-1)]^(3/2) +c

Answer 5

@Callisto i believe the question is quite valid.

Answer 6

wolf's gone nuts
http://www.wolframalpha.com/input/?i=integrate+sqrt%281+%2B+sin%5E2+x%29+sin%28x-1%29+cos%28x-1%29+dx

Answer 7

@apoorvk Hmm... when I messaged the asker , she said :
yes. this is the question. integrate [sqrt of [1 + sin^2(x-1)]]sin(x-1) cos(x-1) exactly what i typed.

Answer 8

I was asking if she had typed the question wrong

Answer 9

*FACEPALM

http://surrealhuntress.files.wordpress.com/2010/06/pic-gun_to_head.jpg

Answer 10

Don't commit suicide!!!

Answer 11

@fatinatikah please edit and correct your question next time if you think you posted it wrong! :)

Answer 12

She thought that was correct :S

Answer 13

LOL ... what's happening??? was the question incorrect??

Answer 14

I won't. From the so many things which are happening since last evening on OS (read ; LaTeX), I'm somehow just stopping myself from doing that. :/

Answer 15

YES @experimentX can you believe it? *sigh*

Answer 16

I was thinking this type of question can't be that difficult... So I asked....

Answer 17

yeah ... it was driving me nuts too ... i guess i was bitten by nutty wolf

Answer 18

And I thought it's OS. It can be anything. So I didn't. @Callisto now you know who's dumber!

(and some wise man said "Ask if you don't know" lol)

Answer 19

Oh... I know...I'm the dumbest :D

Answer 20

No matter where you are, I 'll always be atleast one rung below that :p

Answer 21

save your modesty for someday @Callisto
you really employed beautiful method ...
i would have done let sin x = y
then
sqrt(1+x^2) x dx <--- again substitute ... LOL

Answer 22

''and some wise man said "Ask if you don't know" ''
But from FFM's profile:
''If you have to ask, you'll never know''
Which one is correct?

Answer 23

HAHAHA THANKS EVERYONE. SORRY FOR TYPING THE QUESTION WRONGLY. :)