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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.18 Question 12 Maths Textbook Solution.

Answers (1)

Answer:\sin ^{-1}\left(\frac{\sin x}{2}\right)+c

Hint  \sin x=t

Given: \int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} d x

Explanation:

            \int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} d x            ..............(1)

           Let

           \sin x=t

           \cos x d x=d t                                              (Differentiate w.r.t to t)

   From (1) we have

              \int \frac{d t}{\sqrt{4-t^{2}}}

              =\sin ^{-1}\left(\frac{t}{2}\right)+c \quad\left[\because \int \frac{d t}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}\right]

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

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