Question
Evaluate: $\int_{-1}^1 \frac{1}{ a ^2 e ^x+ b ^2 e ^{-x}} d x$
✓
Answer
$\text { Let } I =\int_{-1}^1 \frac{1}{ a ^2 e ^x+ b ^2 e ^{-x}} d x$
$=\int_{-1}^1 \frac{1}{ a ^2 e ^x+\frac{ b ^2}{ e ^x}} d x$
$=\int_{-1}^1 \frac{ e ^x}{ a ^2\left( e ^x\right)^2+ b ^2} d x$
Put $e ^{ x }= t$
$\therefore e ^{ x } dx = dt$
When $x =-1, t = e ^{-1}$ and when $x =1, t = e$
$\therefore I =\int_{ e ^{-1}}^{ e } \frac{ dt }{ a ^2 t ^2+ b ^2}$
$=\frac{1}{ a ^2} \int_{ e ^{-1}}^{ e } \frac{ dt }{ t ^2+\left(\frac{ b }{ a }\right)^2}$
$=\frac{1}{ a ^2}\left[\frac{1}{\frac{ b }{ a }} \tan ^{-1}\left(\frac{ t }{\frac{ b }{ a }}\right)\right]_{ e ^{-1}}^{ e }$
$=\frac{1}{ ab }\left[\tan ^{-1}\left(\frac{ at }{ b }\right)\right]_{ e ^{-1}}^{ e }$
$\therefore I =\frac{1}{ ab }\left[\tan ^{-1}\left(\frac{ ae }{ b }\right)-\tan ^{-1}\left(\frac{ a }{ be }\right)\right]$
$=\int_{-1}^1 \frac{1}{ a ^2 e ^x+\frac{ b ^2}{ e ^x}} d x$
$=\int_{-1}^1 \frac{ e ^x}{ a ^2\left( e ^x\right)^2+ b ^2} d x$
Put $e ^{ x }= t$
$\therefore e ^{ x } dx = dt$
When $x =-1, t = e ^{-1}$ and when $x =1, t = e$
$\therefore I =\int_{ e ^{-1}}^{ e } \frac{ dt }{ a ^2 t ^2+ b ^2}$
$=\frac{1}{ a ^2} \int_{ e ^{-1}}^{ e } \frac{ dt }{ t ^2+\left(\frac{ b }{ a }\right)^2}$
$=\frac{1}{ a ^2}\left[\frac{1}{\frac{ b }{ a }} \tan ^{-1}\left(\frac{ t }{\frac{ b }{ a }}\right)\right]_{ e ^{-1}}^{ e }$
$=\frac{1}{ ab }\left[\tan ^{-1}\left(\frac{ at }{ b }\right)\right]_{ e ^{-1}}^{ e }$
$\therefore I =\frac{1}{ ab }\left[\tan ^{-1}\left(\frac{ ae }{ b }\right)-\tan ^{-1}\left(\frac{ a }{ be }\right)\right]$
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