Answer:
Let A=∫
0
1
e
lntan
−1
x
⋅sin
(cosx)dx.
We know that e
lnx
=x
Thus, e
ln(tan
x)
=tan
Using this in the above integration, we get :
A=∫
tan
x⋅sin
(cosx)dx
We know that sin
x+cos
x=
2
π
x⋅(
−cos
(cosx))dx
Since, we are integrating from 0 to 1, cos
(cosx)=x
Thus, A=∫
−x)dx
=
∫
xdx−∫
xtan
xdx
Integrating by parts,
A=
([xtan
xdx]
−∫
1+x
dx)−([
x]
2(1+x
)
dx)
((
4
−0)−[
ln(1+x
)]
)−((
×
−
8
ln(2)+
Step-by-step explanation:
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have a good day ahead
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Answers & Comments
Answer:
Let A=∫
0
1
e
lntan
−1
x
⋅sin
−1
(cosx)dx.
We know that e
lnx
=x
Thus, e
ln(tan
−1
x)
=tan
−1
x
Using this in the above integration, we get :
A=∫
0
1
tan
−1
x⋅sin
−1
(cosx)dx
We know that sin
−1
x+cos
−1
x=
2
π
A=∫
0
1
tan
−1
x⋅(
2
π
−cos
−1
(cosx))dx
Since, we are integrating from 0 to 1, cos
−1
(cosx)=x
Thus, A=∫
0
1
tan
−1
x⋅(
2
π
−x)dx
=
2
π
∫
0
1
tan
−1
xdx−∫
0
1
xtan
−1
xdx
Integrating by parts,
A=
2
π
([xtan
−1
xdx]
0
1
−∫
0
1
1+x
2
x
dx)−([
2
x
2
tan
−1
x]
0
1
−∫
0
1
2(1+x
2
)
x
2
dx)
=
2
π
((
4
π
−0)−[
2
1
ln(1+x
2
)]
0
1
)−((
2
1
×
4
π
−0)−[
2
1
−
2
1
tan
−1
x]
0
1
)
=
8
π
2
−
4
π
ln(2)+
2
1
Step-by-step explanation:
hi good morning
have a good day ahead