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Solve the following integral
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\[\int\limits_{}^{}\sqrt{16-x^2}\]
Have you learnt trigo substitution?
we may take \[x=4\sin \theta\] to make it more simplified...
I did that, but my answer isn't coming out to what it should be
\[\int \sqrt{16-(4\sin \theta)^2}d\theta=\int \sqrt{16-16\sin^2\theta} d \theta=4\int\sqrt{1-\sin^2\theta }d\theta =4\int \cos\theta d\theta\\=4(\sin \theta+c) \]
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