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Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 43

Answers (1)

Answer:

        \frac{1}{3}log\left | sin3x \right |-\frac{1}{5}log\left | sin5x \right |+C

Hint:

        sin(A-B)=sinAcosB-cosAsinB

Given:

        \int \! \frac{sin2x}{sin5x\, sin3x}dx

Explanation:

        \int \! \frac{sin(5x-3x)}{sin5x\, sin3x}dx

        =\int \! \frac{sin5x\, cos3x-cos5x\, sin3x}{sin5x\, sin3x}dx

        =\int \! \frac{cos3x}{sin3x}dx-\int \! \frac{cos5x}{sin5x}dx

        =\int \! \cot3xdx-\int \! cot5xdx

        =\frac{1}{3}log\left | sin3x \right |-\frac{1}{5}log\left | sin5x \right |+C

Posted by

Gurleen Kaur

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