Lucille Smitherman

2021-11-18

Evaluate the integral.
$\int \left({e}^{x}-{e}^{-x}\right){\left({e}^{x}+{e}^{-x}\right)}^{3}dx$

Supoilign1964

Let substitute $t={e}^{x}+{e}^{-x}$.
Differentiating both sides, we get
$\frac{dt}{dx}=\frac{d}{dx}\left({e}^{x}+{e}^{-x}\right)$
$\frac{dt}{dx}={e}^{x}-{e}^{-x}$
$dt=\left({e}^{x}-{e}^{-x}\right)dx$
Step 3
Substituting the value in the given integral.
$\int \left({e}^{x}-{e}^{-x}\right){\left({e}^{x}+{e}^{-x}\right)}^{3}dx=\int {t}^{3}dt$
$=\frac{{t}^{4}}{4}+C$
$=\frac{{\left({e}^{x}+{e}^{-x}\right)}^{4}}{4}+C$ [$t={e}^{x}+{e}^{-x}$]
Hence, $\int \left({e}^{x}-{e}^{-x}\right){\left({e}^{x}+{e}^{-x}\right)}^{3}dx=\frac{{\left({e}^{x}+{e}^{-x}\right)}^{4}}{4}+C$

James Obrien

Step 1: Remove parentheses.
$\int \left({e}^{x}-{e}^{-x}\right){\left({e}^{x}+{e}^{-x}\right)}^{3}dx$
Step 2: Use Integration by Substitution.
Let $u={e}^{x}+{e}^{-x},du={e}^{x}-{e}^{-x}dx$
Step 3: Using u and du above, rewrite $\int \left({e}^{x}-{e}^{-x}\right){\left({e}^{x}+{e}^{-x}\right)}^{3}dx$.
$\int {u}^{3}du$
Step 4: Use Power Rule: $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$.
$\frac{{u}^{4}}{4}$
Step 5: Substitute $u={e}^{x}+{e}^{-x}$ back into the original integral.
$\frac{{\left({e}^{x}+{e}^{-x}\right)}^{4}}{4}$
$\frac{{\left({e}^{x}+{e}^{-x}\right)}^{4}}{4}+C$