Question

x
Find Integration, y = x/a®2+x®2​

Answer 1

Answer:

Explanation:

Here,

I

=

∫

x

2

(

a

2

−

x

2

)

3

2

d

x

Let ,

x

=

a

sin

t

⇒

d

x

=

a

cos

t

d

t

∴

a

2

−

x

2

=

a

2

−

a

2

sin

2

t

=

a

2

(

1

−

sin

2

t

)

=

a

2

cos

2

t

So,

I

=

∫

a

2

sin

2

t

(

a

2

cos

2

t

)

3

2

a

cos

t

d

t

=

∫

a

3

sin

2

t

cos

t

a

3

cos

3

t

d

t

=

∫

sin

2

t

cos

2

t

d

t

=

∫

tan

2

t

d

t

=

∫

(

sec

2

t

−

1

)

d

t

=

tan

t

−

t

+

c

=

sin

t

cos

t

−

t

+

c

=

sin

t

√

1

−

sin

2

t

−

t

+

c

Now,

x

=

a

sin

t

⇒

sin

t

=

x

a

and

t

=

sin

−

1

(

x

a

)

Hence,

I

=

x

a

√

1

−

(

x

2

a

2

)

−

sin

−

1

(

x

a

)

+

c