Answer:
a) True
[tex]\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha =0[/tex]
Step-by-step explanation:
Step(i):-
Given that the definite integration
[tex]\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha[/tex]
we know that the trigonometric formula
sin²∝+cos²∝ = 1
cos²∝ = 1-sin²∝
step(ii):-
Now the integration
[tex]\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha = \int\limits^\pi _0 {(\sqrt{cos^{2} \alpha } } \, )d\alpha[/tex]
= [tex]\int\limits^\pi _0 {cos\alpha } \, dx[/tex]
Now, Integrating
[tex]= ( sin\alpha )_{0} ^{\pi }[/tex]
= sin π - sin 0
= 0-0
= 0
Final answer:-
[tex]\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha =0[/tex]
