Question

WHO CAN solve it Please !

Answer 1

Answer:

a) True

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

Step-by-step 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:-

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