Indefinite Integration(11th And 12th > Mathematics ) Questions and Answers
Explanation:-
Answer: Option A. -> ∑9n=1tannxn:
A
We have In=∫tannxdx=∫tann−2x(sec2x−1)dx=tann−1xn−1−In−2i.e.In−2+In=tann−1xn−1(n≥2)
Thus, we have
I0+I1+2(I2+....+I8)+I9+I10=(I0+I2)+(I1+I3)+(I2+I4)+(I3+I5)+(I4+I6)+(I5+I7)+(I6+I8)+(I7+I9)+(I8+I10).=∑10n=2tann−1xn−1=∑9n=1tannxn
Explanation:-
Answer: Option B. -> log|sec(x2+12)|+C:
B
We have, ∫x√2sin(x2+1)−sin2(x2+1)2sin(x2+1)+sin2(x2+1)dx =∫x√2sin(x2+1)−2sin(x2+1)cos(x2+1)2sin(x2+1)+2sin(x2+1)cos(x2+1)dx =∫x√1−cos(x2+1)1+cos(x2+1)dx =∫xtan(x2+12)dx =∫tan(x2+12)d(x2+12) =log|sec(x2+12)|+C
Explanation:-
Answer: Option A. -> 1√2tan−1(1√2tan 2x)+C:
A
∫dxsin4 x +cos4 x=∫dx(sin2 x+cos2 x)2−2 sin2 x cos2 x=∫2 dx2−sin2 2x=∫2 sec2 2x2 sec2 2x−tan2 2xdx=2∫sec2 2x2+tan2 2xdx=∫dt(√2)2+t2[putting tan 2x=t⇒2 sec2 2x dx=dt]=1√2tan−1(t√2)+C=1√2tan−1(1√2tan 2x)+C