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  • Integrate the functions in Exercises 1 to 24. 9

Integrate the functions in Exercises 1 to 24.

    Q9.    \frac{\cos x}{\sqrt{4 - \sin^2 x}}

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We have the given integral

                                                    I\ =\ \frac{\cos x}{\sqrt{4 - \sin^2 x}}

Assume                              \sin x = t\ \Rightarrow \cos x dx = dt

So, this substitution gives,

                                          \int \frac{\cos x}{\sqrt{4 - \sin^2 x}}\ =\ \int \frac{dt}{\sqrt{(2)^2 - (t)^2}}

                                                                             =\ \sin^{-1}\frac{t}{2}\ +\ C

or                                                                           =\ \sin^{-1}\left ( \frac{\sin x}{2} \right )\ +\ C

Posted by

Devendra Khairwa

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