Interestingly, this is the answer I kept receiving, yet the solutions manual shows: 1/4[(tan)^4 x] - [(tan)^2 x] + ln |(sec)^2 x| Did they misprint the solution? Where did the 1/2 before the second term disappear to?
4:03 i did secx=t, secx.tanx.dx=dt instead of yours and i found "sec^4(x)/4 - sec^2(x) - ln(cosx) + c" so am i missing something or both of this are true? edit: i searched a bit and yes both of this are true
Thank you so much. This was incredibly helpful. Peace be upon you.
Interestingly, this is the answer I kept receiving, yet the solutions manual shows:
1/4[(tan)^4 x] - [(tan)^2 x] + ln |(sec)^2 x|
Did they misprint the solution? Where did the 1/2 before the second term disappear to?
It's a indefinite intigration. There can be different solution or you can solve more
Extremely good
trig reduction formulas make this one easier
1/(n-1) tan^(n-1)(x) - integral of (tan^(n-2)(x)dx)
It was very helpful for me and hope for other's
Thany u so much sir..🎉
4:03 i did secx=t, secx.tanx.dx=dt instead of yours and i found "sec^4(x)/4 - sec^2(x) - ln(cosx) + c" so am i missing something or both of this are true?
edit: i searched a bit and yes both of this are true
Thanks for the update
I found the same as yours too
Thank you sir,
I thank God of u sir,l wish to see you physically one day.
Amazing
There's still tan3x why isn't tan2x substituted for it
What lol
Calculus 2,o salute.so difficult
Mst
A video on underroot tanx plz
lemme guess reduction formula
there is an easier way :1- Separate tan^5x to tan^4(tanx)
2-and then use the identity :sec^2x-1=tanx
3-let secx=u ...