37 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.When u input ABC and actually calculate the numerator it doesn't come out to be 1
Ayeee finally something math I can do
Lol
Check my working
Hmm I will wait
It works but it will be 1-t²
And not 1+t²
In the denominator
I took -cosx as t
Bro came in and observed before me 😭🙏
So t² is cos² and not -cos²
AH
😭
I'd have done by parts here though
Idk why
Feels like it
i avoid byparts when its 1/f(x)
more prone to error
Hmm I see
Vaise your thing works well too
By parts was the first hunch I had
hmmm
well
my ans still doesnt match
the substitution t = tan u/2 is easier though here
isnt that a standard form?
i dont recall 1/1-tan^2x form
2t/1+t2 for sin u
And
1-t2/1+t2 for cos u
Sorry लोल
Edited
This should be easier to read
$\frac{-dt}{(1-t^{2})(2+t)}=\frac{dt}{2(2+t)}(\frac{1}{t-1}-\frac{1}{t+1})=(\frac{1}{6(2+t)}-\frac{1}{6(t-1)}-\frac{1}{2(t+1)}+\frac{1}{2(t+2)})=\frac{1}{3(t+2)}-\frac{1}{6(t-1)}-\frac{1}{2(t+1)}$
SirLancelotDuLac
So, we would get $\frac{(ln(\abs{\frac{(t+2)^{2}}{(t-1)(t+1)^{3}}}))}{6}$
Where t=cos(x)
SirLancelotDuLac
thats more work
i just messed up partal fractions (twice 💀 )
there isnt -dt but yeah got my mistake
+Solved @SirLancelotDuLac
Ye I took t=cos(t) rather than its negative.
Also rip bot.
lol
:(
Aadat ki baat hai.
Answer to wahi :) log of something na.
yeahh
Options dekh ke you can figure. Partial fractions me wahi risk hai
If you mess numbers answer completely alag aayega.
adv doesnt ask indefinite
लोल this is for main?
Then it doesn’t matter.
aur jo mains mein aate they usually go like f(x) = integral . f(1)= something find f(something else)
Do they ask things like this ask kal?
Right.
so options dekh ke guess kar pana is a rare ocurrence
+solved @SirLancelotDuLac
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