Question

Evaluate the integral:
$\color{darkviolet}\boxed{ \tt \ \displaystyle \gg\gg\tt∫cos x \sqrt{ \left(9 - sin {}^{2} x \right)} dx \ll\ll}$
Answer with proper explanation.

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Answer 1

Question:-

Evaluate the integral:

$\sf \large \int cos x \sqrt{ \left(9 - sin {}^{2} x \right)} dx .$

Given:-

$\sf \large \int cos x \sqrt{ \left(9 - sin {}^{2} x \right)} dx .$

To Find:-

• To Evaluate the Given integral.

Solution:-

$\sf \large I = \sf \large \int cos x \sqrt{ \left(9 - sin {}^{2} x \right)} dx .$

Put, sin x = t, then cos x dx = dt.

$\sf \large \therefore I = \int \sqrt{(9 - t {}^{2}) } dt$

We have that,

$\sf \large \int \sqrt{a {}^{2} - x {}^{2} )} dx = \frac{1}{2} \bigg \{x \sqrt{(a {}^{2} - x {}^{2}) } + a {}^{2} \: sin {}^{ - 1} \frac{x}{a} \bigg \}.$

$\sf \large I = \int \sqrt{(9 - t) {}^{2} } dt = \frac{1}{2} \bigg \{t \sqrt{(9 - t {}^{2} )} + 9 \: sin {}^{ - 1} \: \frac{t}{3} \bigg \} + c.$

$\sf \large = \frac{1}{2 } \bigg \{sin \: x \sqrt{(9 - sin {}^{2} x)} + 9 \: sin {}^{ - 1} \bigg( \frac{sin \: x}{3} \bigg) \bigg \} + c.$

Answer:-

$\sf \large \color{violet} \sf \large \int cos x \sqrt{ \left(9 - sin {}^{2} x \right)} dx \sf \large = \frac{1}{2 } \bigg \{sin \: x \sqrt{(9 - sin {}^{2} x)} + 9 \: sin {}^{ - 1} \bigg( \frac{sin \: x}{3} \bigg) \bigg \} + c.$

Hope you have satisfied. ⚘