Evaluate $\int_{\log \sqrt{2}}^{\log \sqrt{3}} \frac{1}{\left(e^{x}+e^{-x}\right)\left(e^{x}-e^{-x}\right)} d x$

CBSE Class 12 · 2023 · Definite IntegralsShort Answer

Evaluate

\int_{l o g 2}^{l o g 3} \frac{1}{( e ^{x} + e ^{- x} ) ( e ^{x} - e ^{- x} )} d x

Answer: $\frac{1}{4} lo g \frac{3}{2}$

First simplify the denominator:

(e^{x} + e^{- x}) (e^{x} - e^{- x}) = e^{2 x} - e^{- 2 x}

.
Put

t = e^{2 x} \Rightarrow d t = 2 e^{2 x} d x \Rightarrow d x = \frac{d t}{2 t}

, and

e^{2 x} - e^{- 2 x} = t - \frac{1}{t} = \frac{t ^{2} - 1}{t}

.
Limits:

x = lo g 2 \Rightarrow t = e^{l o g 2} = 2

;

x = lo g 3 \Rightarrow t = 3

I = \int_{2}^{3} \frac{t}{t ^{2} - 1} \cdot \frac{d t}{2 t} = \frac{1}{2} \int_{2}^{3} \frac{d t}{t ^{2} - 1} = \frac{1}{4} [lo g \frac{t - 1}{t + 1}]_{2}^{3}

= \frac{1}{4} [lo g \frac{2}{4} - lo g \frac{1}{3}] = \frac{1}{4} [lo g \frac{1}{2} + lo g 3] = \frac{1}{4} lo g \frac{3}{2}

CBSE Class 12 Board · 2023 · Subjective