Question

SOLVE THIS
INTEGRAL???​

Answer 1

Step-by-step explanation:

choose the correct sign (+,-,× or =) in the blank space.

8_6_7_2

Answer 2

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