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Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 48 maths

Answers (1)

Answer:

        cot\, \theta log\left |\frac{cos\, x}{cos(x+\theta )} \right |+C

Hint:

        tan2x=\frac{2tan\, x}{1-tan^{2}x}

Given:

        \int \! [1+tan\, x\, tan\left | x+\theta \right |]dx                        ...(1)

Explanation:

        tan\, \theta =tan[x+\theta -x]=\frac{tan(x+\theta )-tan\, x}{1+tan\, x\, tan(x+\theta )}

        1+tan\, x\, tan(x+\theta )=cot\, \theta [tan(x+\theta )-tan\, x]

Put in (1)

        \int \! cot\, \theta [tan(x+\theta )-tan\, x]dx

        =cot\, \theta \int \! tan(x+\theta )dx-cot\, \theta \int \! tan\, xdx

        =cot\, \theta [-log\left | cos(x+\theta ) \right |+log\left | cos\, x \right |]+C

        =cot\, \theta log\left |\frac{cos\, x}{cos(x+\theta )} \right |+C

Posted by

Gurleen Kaur

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