Question

i got stuck lol...

\(\Huge \int \frac{\cos x dx}{(\sin x)^3 + \sin x}\)

Answer 1

Take sin x =t
dt = cosx dx

\[\int\limits_{}^{}dt/ t(1+t^2)\]

Apply partial fraction. . I hope you know what is partal fraction

Answer 2

\[\int \frac{du}{u^3+u}\]

Answer 3

ahh partial fractions of course...but for the sake of it..would you mind showing it?

Answer 4

just a wild question...is trig sub applicable in u^2 + u?

Answer 5

\[\frac{a}{x}+\frac{bx+c}{x^2+1}=\frac{1}{x(x^2+1)}\]
\[a(x^2+1)+x(bx+c)=1\]
let \(x=0\) get \(a=1\)

Answer 6

In any case.

We can write

\[1/(t)(1+t^2) = A/t + (Bx+C)/1+t^2\]
\[1=A(1+t^2) +(Bx+C)(t)\]
Where A,B,C are constants/
Now substitute different values of x to get A, B and C
Finally integrate :)

Answer 7

and since
\[a(x^2+1)+x(bx+c)=1\]
\[x^2+1+x(bx+c)=1\]
\[x^2+bx^2=(1+b)x^2\implies b=-1\] \]

Answer 8

how did c become zero?

Answer 9

and \(cx\implies c=0\) s you get
\[\int\frac{du}{u}-\int \frac{udu}{u^2+1}\]

Answer 10

c is zero because if you expand the right left hand side you end up with a \(cx\) term, but on the right hand side there is no x term. that means \(c=0\)

Answer 11

ahh i see..silly me =))

Answer 12

so from that im seeing ln u - [ln (u^2 + 1)]/2

Answer 13

just a wild question...is trig sub applicable in u^2 + u?

Yes. I think this method will be faster/

\[\int\limits_{}^{}dt/(t)(1+t^2)\]

Substitute t=tanx

\[\int\limits_{}^{} \sec^2x dx/ \tan x(1+\tan^2x) \]

Since 1+tan^2 (x) = sec^2 (x)

Therefore sec^2(x) in numerator and denominator gets cancelled.

So we are left with
\[\int\limits_{}^{}\cot x dx = \log \left| sinx \right| + c\]

Answer 14

Short and sweet :)

Answer 15

Final answer is
\[\log \left| \sin(\tan^{-1} t) \right|+C\]

Answer 16

but wolfram gave a different answer :/

Answer 17

i got ln (sinx) - 1/2 ln (sin^2 x + 1)

Answer 18

We always have many possible answers to the same integration question. They all differ by constant. can you point out any mistake in my method ??

Answer 19

well in the trig substitution we were the same until log |sin x| i think what x is...we're different haha

Answer 20

no..we're the same all through out...in log|sin theta| you shouldve gotten

log|(t)/sqrt(t^2 + 1))

Answer 21

log t - 1/2 log | t^2 + 1|

change back to sinx

log |sin x| - 1/2 log{sin^2 x + 1|

Answer 22

you messed up when you reused x :P when it was in the original problem already

Answer 23

Yes. I agree @lgbasallote . YOu are right I messed it up taking x

Answer 24

though what i dont get is why the second term of wolfram is -1/2 ln (3 - cos(2x)) can you provide me with an answer??

Answer 25

ok. Which method-- partial fraction or the trig substitution?

Answer 26

partial

Answer 27

ok. :) GIve me 5 minutes. I am very slow in typing

Answer 28

\[I= \int\limits_{}^{} dt/t - \int\limits_{}^{} tdt/1+t^2\]
Let p=1+t^2
dp/2 = t* dt

\[I=\log t - (1/2) \int\log dp/p\]
\[I=\log t -(1/2) \log p + C\]
substituting p= 1+t^2
\[I=\log t- (1/2)(\log (1+t^2)) +C\]
substituting t= sin x
\[I = \log (\sin x) - (1/2)(\log(1+(\sin x)^2)) +C\]

Answer 29

Now We know that cos 2x = 1 - 2 (sinx)^2

Therefore
(sinx)^2 = (1 - cos 2x)/2

Answer 30

\[I = \log (\sin x) - (1/2)(\log(1+(1-\cos 2x)/2)\]
\[I = \log (\sin x) - (1/2)(\log(2+1-\cos 2x)/2)\]
\[I = \log (\sin x) - (1/2)(\log((3-\cos 2x)/2))\]

Answer 31

@lgbasallote , Hope you got it :)

Answer 32

lol I took 20 minutes to type this. My bad :(

Answer 33

@Callisto , beat this record. 20 minutes to type half solution to a integral problem :)

Answer 34

@shivam_bhalla lol I'll try if I have 30 minutes :P Perhaps in your question ? lol

Answer 35

lol it seems they were just the same :P wolfram just did some crazy thing..20 mins wasted >:)))

Answer 36

Seriously :P I told you. Most of the answers are simply manipulating among trig identities or differing by a constant

Answer 37

But I always don't know which trigo identities to use :(

Answer 38

@Callisto , it only comes by practice and the experience gained out of practicing :)

Answer 39

That's why I'm trying to remember all the cases :(