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  • Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 28 maths

Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 28 maths

Answers (1)

Answer:

        \frac{1}{sin(b-a)}[log(\frac{sec(x+b)}{sec(x+a)})]+C

Hint:

        M\! ultipl\! y\; and\; divide\; by\; sin(b-a)

Given:

        \int \! \frac{1}{cos(x+a)cos(x+b)}dx                    ......(1)

Explanation:

M\! ultipl\! y\; and\; divide\; by\; sin(b-a)

From (1)

        =\frac{1}{sin(b-a)}\int \! \frac{sin(b-a)}{cos(x+a)cos(x+b)}dx

        =\frac{1}{sin(b-a)}\int \! \frac{sin[(x+b)-(x+a)]}{cos(x+a)cos(x+b)}dx        Add and subtract x in b-a

        =\frac{1}{sin(b-a)}\int \! \frac{sin(x+b)cos(x+a)-sin(x+a)cos(x+b)}{cos(x+a)cos(x+b)}dx

        =\frac{1}{sin(b-a)}[\int \! \frac{sin(x+b)}{cos(x+b)}dx-\int \! \frac{sin(x+a)}{cos(x+a)}dx]

        =\frac{1}{sin(b-a)}[\int \! tan(x+b)dx-\int \! tan(x-a)dx]

        =\frac{1}{sin(b-a)}[log\left | sec(x+b) \right |-log\left | sec(x+a) \right |]+C

        =\frac{1}{sin(b-a)}[log\left | \frac{sec(x+b)}{sec(x+a)} \right |]+C                [log\, a-log\, b=log\frac{a}{b}]

Posted by

Gurleen Kaur

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