A.
∫cotxdx=?
Let I=∫cotxdx
I=∫cosxsinxdx−(1)
Put sinx=t−(2)
⇒cosxdx=dt−(3)
Put (3) in (1)
I=∫dtt=log|t|+c
I=log|sinx|+c
∫cotxdx=log|sinx|+c
B.
∫tanxdx=?
Let I=∫tanxdx
I=∫sinxcosxdx−(1)
Put cosx=t−(2)
⇒−sinxdx=dt−(3)
Put (3) in (1)
I=∫dtt=−log|t|+c
I=−log|cosx|+c
I=log∣∣∣1cosx∣∣∣+c
I=log|secx|+c
∫tanxdx=log|secx|+c
C.
∫11+cos2xdx=?
Using, cos2x=2cos2x−1
I=∫11+cos2xdx=∫11+2cos2x−1dx
I=∫12cos2xdx
=12∫sec2xdx
Using ∫sec2xdx=tanx
I=tanx2+c
D.
∫(1+tan2x)dx=?
As, we know that 1+tan2x=sec2x
∴∫(1+tan2x)dx=∫(sec2x)dx
Using, ddxtanx=sec2x
We have,
∫(sec2x)dx=tanx+c
∫(1+tan2x)dx=tanx+c
E.
∫cscxdx=?
I=∫cscxdx
=∫cscx(cscx+cotx)cscx+cotxdx−(1)
cscx+cotx=t−(2)
−cscx(cscx+cotx)dx=dt
Put (3) in (1)
I=∫−dtt
I=−log|t|+c=−log|cscx+cotx|+c
−log|cscx+cotx|+c=−log∣∣∣1sinx+cosxsinx∣∣∣+c
=−log∣∣∣1+cosxsinx∣∣∣+c
=−log∣∣
∣
∣
∣∣1+(2cos2x2−1)sinx∣∣
∣
∣
∣∣+c
=−log∣∣
∣
∣∣2cos2x22sinx2cosx2∣∣
∣
∣∣+c
=−log∣∣
∣
∣∣2cosx2cosx22sinx2cosx2∣∣
∣
∣∣+c
=−log∣∣
∣
∣∣cosx2sinx2∣∣
∣
∣∣+c
=−log∣∣∣cotx2∣∣∣+c
=log∣∣
∣
∣∣1cotx2∣∣
∣
∣∣+c
=log∣∣∣tanx2∣∣∣+c