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  • Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 30

Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 30

Answers (1)

Answer:

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

Hint:

        U\! se\; cos\, A-cos\, B \: and\: sin\, A-sin\, B f\! ormula

Given:

        \int \! \frac{cos4x-cos2x}{sin4x-sin2x}dx

Explanation:

        \int \frac{-2 \sin \left(\frac{4 x-2 x}{2}\right) \sin \left(\frac{4 x+2 x}{2}\right)}{2 \cos \left(\frac{4 x+2 x}{2}\right) \sin \left(\frac{4 x-2 x}{2}\right)} d x \quad\left[\begin{array}{l} \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \\ \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \end{array}\right]

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

        =-\int \! \tan\, 3xdx

        =-[\frac{-log\left | cos\, 3x \right |}{3}]+C

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

Posted by

Gurleen Kaur

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