Answer:
Step-by-step explanation:
1. Given the integral function [tex]\int\limits {\sqrt{a^{2} -x^{2} } } \, dx[/tex], using trigonometric substitution, the substitution that will be most helpful in this case is substituting x as [tex]asin \theta[/tex] i.e [tex]x = a sin\theta[/tex].
All integrals in the form [tex]\int\limits {\sqrt{a^{2} -x^{2} } } \, dx[/tex] are always evaluated using the substitute given where 'a' is any constant.
From the given integral, [tex]\int\limits {7\sqrt{49-x^{2} } } \, dx = \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx[/tex] where a = 7 in this case.
The substitute will therefore be [tex]x = 7 sin\theta[/tex]
2.) Given [tex]x = 7 sin\theta[/tex]
[tex]\frac{dx}{d \theta} = 7cos \theta[/tex]
cross multiplying
[tex]dx = 7cos\theta d\theta[/tex]
3.) Rewriting the given integral using the substiution will result into;
[tex]\int\limits {7\sqrt{49-x^{2} } } \, dx \\= \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -(7sin\theta)^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -49sin^{2}\theta } } \, dx\\= \int\limits {7\sqrt{49(1-sin^{2}\theta)} } } \, dx\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, dx\\since\ dx = 7cos\theta d\theta\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, 7cos\theta d\theta\\= \int\limits {7\{7(cos\theta)} }}} \, 7cos\theta d\theta\\[/tex]
[tex]= \int\limits343 cos^{2} \theta \, d\theta[/tex]
