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  • Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 6

Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 6

Answers (1)

Answer:

        tan^{-1}(e^{x})+C

Hint:

Use substitution method as well as special integration formula to solve this type of problem

Given:

\int \frac{1}{e^{x}+e^{-x}}dx

Solution:

Let\: \:I= \int \frac{1}{e^{x}+e^{-x}}dx=\int \frac{1}{e^{x}+\frac{1}{e^{x}}}dx

            \begin{aligned} &=\int \frac{1}{\frac{e^{2 x}+1}{e^{x}}} d x \\ &=\int \frac{e^{x}}{e^{2 x}+1} d x \end{aligned}

Put\: \: e^{x}=t\Rightarrow e^{x}dx=dt

Then\: \: \begin{aligned} I=\int \frac{1}{t^{2}+1} d t \end{aligned}

                    \begin{array}{ll} =\tan ^{-1}(t)+C & {\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]} \\\\ =\tan ^{-1}\left(e^{x}\right)+C & {\left[\because e^{x}=t\right]} \end{array}

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