what is the derivative of√cot^2√tanx
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Answer:
MATHS
Evaluate :
∫
0
π/2
(
tanx
+
cotx
)dx
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ANSWER
I=∫
0
2
π
(
tanx
+
cotx
)dx
I=∫
0
2
π
(
cosx
sinx
+
sinx
cosx
)dx
I=∫
0
2
π
(
sinxcosx
sinx+cosx
)dx
Letz=sinx−cosx,dz=(cosx+sinx)dx
z
2
=sin
2
x+cos
2
x−2sinxcosx
z
2
=1−2sinxcosx
2sinxcosx=1−z
2
sinxcosx=
2
1−z
2
When,x=0,t=−1;x=
2
π
,t=1
I=∫
−1
1
1−t
2
2
dz
I=[
2
sin
−1
t]
−1
1
I=
2
(
2
π
−
2
3π
)
I=−
2
π
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