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  • Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 6 maths textbook solution.

Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 6 maths textbook solution.

Answers (1)

\tan ^{-1}e-\frac{\pi}{4}

Hint: We use indefinite formula then put limits to solve this integral.

Given: \int_{0}^{1}\frac{e^x}{1+e^{2x}}dx

Solution: \int_{0}^{1}\frac{e^x}{1+e^{2x}}dx

Put e^x=t

e^x dx=dt

When x=0  then t=1  and when x=1  then t=e

\begin{aligned} &I=\int_{1}^{e} \frac{1}{1+t^{2}} d t \\ &=\left[\tan ^{-1} t\right]_{1}^{e} \\ &=\tan ^{-1} e-\tan ^{-1} 1\; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} 1=\frac{\pi}{4}\right] \end{aligned}

=\tan ^{-1}e-\frac{\pi}{4}

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