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Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 20 maths

Answers (1)

Answer:

        \frac{1}{3}log\left | 2+3\, sin\, x \right |+C

Hint:

        \frac{\mathrm{d} }{\mathrm{d} x}(a+b\, sin\, x)=cos\, x

Given:

        \int \! \frac{cos\, x}{2+3\, sin\, x}dx                    .....(1)

Explanation:

Let

        2+3\, sin\, x=t

        3\, cos\, x\, dx=dt

        cos\, x\, dx=\frac{dt}{3}

Put in (1)

        \int \! \frac{dt}{3t}=\frac{1}{3}\int \! \frac{dt}{t}

        =\frac{1}{3}log\left | t \right |+C

        =\frac{1}{3}log\left | 2+3\, sin\, x\right |+C

Posted by

Gurleen Kaur

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