Question

integrate dx/1+2sin^x from 0 to pie

Answer 1

@ganeshie8

Answer 2

@genny7

Answer 3

Do you mean\[\int\frac{dx}{1+2\sin^2x}~~?\]

Answer 4

its definite integration from 0 to pie

Answer 5

Yes I get that (but omitted it), but the integrand you provided doesn't make sense.

Answer 6

i didnt understand

Answer 7

What do you mean by \(\sin^x\)? What is the argument?

Answer 8

its sin^2x

Answer 9

So it is
\[\int_0^\pi\frac{dx}{1+2\sin^2x}~~?\]

Answer 10

yes

Answer 11

divide top and bottom by cos^2x

Answer 12

\[\large \large \int \dfrac{dx}{1+2\sin^2x} = \int \dfrac{\sec^2x~dx}{1+3\tan^2x}\]

Answer 13

next substitution is quite obvious..

Answer 14

Another way to do it is dividing by \(\sin^2x\) in place of \(\cos^2x\). Same process in the end.

Answer 15

ok thanks

Answer 16

how limit will change

Answer 17

Ahh looks thats going to be tricky
we need to split the integral i guess

Answer 18

how tell me

Answer 19

i m not getting

Answer 20

@sidsiddhartha

Answer 21

@aum

Answer 22

dividing by sin^2x and substituting cotx works out without complications as cotx is continuous in (0,pi)

but dividing by cos^2x and tanx substitution also must work if we split the integral, right ?

Answer 23

is it an odd function

Answer 24

not sure if i am making sense, im bit confused hmm

Answer 25

This is an even function. So the integral could be 2 * [0, pi/2].

Answer 26

ok

Answer 27

oops sorry. I read the limits wrong.

Answer 28

ok

Answer 29

going for lunch brb, thanks aum :)

Answer 30

someone please help me

Answer 31

\[\begin{align*}\int\frac{dx}{1+2\sin^2x}&=\int\frac{\csc^2x}{\csc^2x+2}~dx\\\\
&=\int\frac{\csc^2x}{(1+\cot^2x)+2}~dx\\\\
&=\int\frac{\csc^2x}{\cot^2x+3}~dx\\\\
&=-\int\frac{du}{u^2+3}&\text{setting }u=\cot x\\&&\text{so }-du=\csc^2x~dx\\\\
&=-\int\frac{\sqrt3\sec^2t}{(\sqrt3\tan t)^2+3}~dt&\text{setting }u=\sqrt 3\tan t\\
&&\text{so }du=\sqrt 3\sec^2t~dt\\\\
&=-\int\frac{\sqrt3\sec^2t}{3\tan^2 t+3}~dt\\\\
&=-\frac{\sqrt{3}}{3}\int\frac{\sec^2t}{\tan^2 t+1}~dt\\\\
&=-\frac{\sqrt{3}}{3}\int dt
\end{align*}\]
Integrate and back-substitute.

Answer 32

i got the answer.by another method just now

Answer 33

looks you need to use some symmetry
http://math.stackexchange.com/questions/958920/evaluate-int-limits-0-pi-dfracdx12-sin2x

whats your other method ?

Answer 34

thanks ganeshie bt i got the answer

Answer 35

yeah i am stuck at understanding the bounds because F(pi) - F(0) gives you 0.

may i knw your other method ?

Answer 36

i break the integation in two parts from o to pie/2 and pie/2 to pie

Answer 37

i did this when i got sec^2dx/1+3tan^2x

Answer 38

you will get 0 right ?

Answer 39

F(pi/2) - F(0) + F(pi) - F(pi/2)

F(pi) - F(0)

0

Answer 40

no pie/root 3

Answer 41

splitting is giving me 0. see above

Answer 42

limit will change from0 to infinity and -infnity to 0

Answer 43

you're right ! splitting will do (: