Question

Integration

Answer 1

oh the suspense...

Answer 2

\[\int\limits_{\pi/10}^{2\pi/5}(\sin x) /(\sin x + \cos x)\]

Answer 3

lol

Answer 4

not an easy one... unless I'm missing a trick

Answer 5

im was thinkin multiply top and bottom by sinx-cosx though i dont see how that helps

Answer 6

solving these stuff really gives me the creeps!!!

Answer 7

why not do what bahrom suggests?

Answer 8

Bahrom is right. We shall get cos2x in the denominator. and above simply sin^2x-sinxcosx
sin^2x=(1-cos2x)/2
sinxcosx=sin2x/2
Now, it should be possible

Answer 9

i always have good ideas :D, i just dont realize it haha.

Answer 10

Yea and i am not modest ;)

Answer 11

even we can use the subsitution u=tan(x/2)

Answer 12

^that is wolframs suggestion and it gets pretty crazy the way that goes

Answer 13

u get( - cos2x ) at denomintor

Answer 14

yeah i looked at that too damned long

Answer 15

x/2=tan^-1(u)
x=2tan^-1(u)
dx=(2)/(1+u^2)du....

Answer 16

Multiply by (cosx-sinx) not (sinx-cosx)

Answer 17

maybe something along the the lines of
\[sin(x)+\cos(x)=\sqrt{2}\sin(x+\frac{\pi}{4})\]?

Answer 18

is the answer 0.5[x+(sec^2 x)/2] as per Bahroms method???

Answer 19

not quite according to wolf

Answer 20

close...

Answer 21

Check my work,
\[\frac{\sin^2x-sinxcosx}{\sin^2x-\cos^2x}\]
\[\frac{\sin^2x-sinxcosx}{\cos2x}\]
\[\huge \frac{(-\frac{\cos2x-1}{2})-sinxcosx}{\cos2x}\]
\[\frac{1-\cos2x-2sinxcosx}{\cos2x}\]
\[\sec2x-1-\tan2x\]

Answer 22

it'll be (-cos2x) in the denominator,

Answer 23

yeah

Answer 24

mani jah method gives
\[\int \frac{\sin(x)(\cos(x)-\sin(x))}{\cos(2x)}dx\]
\[\int\frac{\sin(x)\cos(x)}{\cos(2x)}dx-\int\frac{sin^2(x)}{\cos(2x)}dx\]
\[\frac{1}{2}\int\frac{\sin(2x)}{\cos(2x)}dx-\int \frac{\sin^2(x)}{\cos(2x)}dx\] now first one is easy, second one not sure about

Answer 25

we can turn the second into\[\int{\sin^2x\over2\sin^2x-1}dx\]and um... that's where I feel like I should know what to do...

Answer 26

what is the integration of sec 2x

Answer 27

@satellite73, you can also use sin^2x=(1-cos2x)/2

Answer 28

integral of secx is a totally different question; let's keep it one at a time
(to do it multiply by sec+tan/sec+tan)

Answer 29

As per mani method i am getting
\[0.5\int\limits_{}^{}[\tan 2x+1-\sec 2x]\]

Answer 30

yeah that is correct

Answer 31

I'm too lazy to work out the answer myself right now. Just put the limits now, and see if you get the answer

Answer 32

so i nee the integral of sec 2x @mani

Answer 33

I just told you how Sarkar

Answer 34

i didnt get it could you please,explain!!!

Answer 35

\[\int\sec xdx=\int\sec x\cdot{\sec x+\tan x\over\sec x+\tan x}dx\]\[=\int{\sec^2x+\sec x\tan x\over\sec x+\tan x}dx\]\[u=\sec x+\tan x\]

Answer 36

See no. 73. http://integral-table.com/old/IntegralTable.pdf

Answer 37

okay!!!