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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 13 maths textbook solution

Answers (1)

Answer:

        \frac{1}{3}log\left | 3sec\; x+5 \right |C

Himt:

        \int \! \frac{1}{x}dx=log\left | x \right |+C

Given:

        \int \! \frac{sec\: x\: tan\: x}{3sec\: x+5}dx

Explanation:

Let,

        3sec\: x+5=t

        3sec\: x \: tan \: xdx=dt

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

                                    =\frac{1}{3}log \: t+C

                                    =\frac{1}{3}log \: \left | 3sec\: x+5 \right |+C

 

 

 

Posted by

Gurleen Kaur

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