Question

WHO CAN solve it Please !

Answer 1

a) True

`\int\limits^\pi _ {0} \,(\sqrt{1-sin^(2)\alpha } )d\alpha =0`

Explanation:

Step(i):-

Given that the definite integration

`\int\limits^\pi _ {0} \,(\sqrt{1-sin^(2)\alpha } )d\alpha`

we know that the trigonometric formula

sin²∝+cos²∝ = 1

cos²∝ = 1-sin²∝

step(ii):-

Now the integration

`\int\limits^\pi _ {0} \,(\sqrt{1-sin^(2)\alpha } )d\alpha = \int\limits^\pi _0 {(\sqrt{cos^(2) \alpha } } \, )d\alpha`

=
`\int\limits^\pi _0 {cos\alpha } \, dx`

Now, Integrating

`= ( sin\alpha )_(0) ^(\pi )`

= sin π - sin 0

= 0-0

= 0

Final answer:-

`\int\limits^\pi _ {0} \,(\sqrt{1-sin^(2)\alpha } )d\alpha =0`