1. Given the integral function $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ , using trigonometric substitution, the substitution that will be most helpful in this case is substituting x as $asin \theta$ i.e $x = a sin\theta$ .

All integrals in the form $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ are always evaluated using the substitute given where 'a' is any constant.

From the given integral, $\int\limits {7\sqrt{49-x^{2} } } \, dx = \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx$ where a = 7 in this case.

The substitute will therefore be $x = 7 sin\theta$

2.) Given $x = 7 sin\theta$

$\frac{dx}{d \theta} = 7cos \theta$

cross multiplying

$dx = 7cos\theta d\theta$

3.) Rewriting the given integral using the substiution will result into;

$\int\limits {7\sqrt{49-x^{2} } } \, dx \\= \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -(7sin\theta)^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -49sin^{2}\theta } } \, dx\\= \int\limits {7\sqrt{49(1-sin^{2}\theta)} } } \, dx\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, dx\\since\ dx = 7cos\theta d\theta\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, 7cos\theta d\theta\\= \int\limits {7\{7(cos\theta)} }}} \, 7cos\theta d\theta\\$

$= \int\limits343 cos^{2} \theta \, d\theta$

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4 years ago