6 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.I feel like x=sin(theta) is a better substitution
'Cuz dx=cos(theta)d(theta) and then you just have integral of sec^2(2theta)d(theta)
Which is tan(2theta)/2 with theta from 0 to pi/6 which gives you sqrt(3)/2
Umm can you solve it ?
Manipulation
Thoda sa pen chalayo 😃
Cool tip though: when see these difference of squares under the square roots, it usually gets easier with trigonometric substitution
Sinx, tanx or something else..... That depends
It's based solely on just two identities
sin square plus cos square
And sec square minus tan square
I did that above though?
But anyhoo, substituting x=sin(t) dx=cost.dt So, the thing above is cost.dt/(1-2sin²t)cost which is sec²2t
And the rest is as above