Question

37–56. Integrals Evaluate each integral.∫ sinh x / (1 + cosh x) dx

Accepted Answer

Recall the definitions and identities for hyperbolic functions: $$\sinh x = \frac{e^x - e$$^{-x}$$}{2}$$ and $$\cosh x = \frac{e^x + e$$^{-x}$$}{2}. $$Also, remember the identity $$1 + \cosh x = 2 \$$cosh^2$$ \left(\frac{x}{2}ight). $$Rewrite the integral using the identity for the denominator: $$\int \frac{\sinh x}{1 + \cosh x} \, dx = \int \frac{\sinh x}{2 \$$cosh^2$$ \left(\frac{x}{2}ight)} \, dx. $$Express $$\sinh x in $$terms of $$\sinh \frac{x}{2}$$ and $$\cosh \frac{x}{2}$$ using the double-angle formula: $$\sinh x = 2 \sinh \left(\frac{x}{2}ight) \cosh \left(\frac{x}{2}ight). $$Substitute this expression into the integral to get $$\int \frac{2 \sinh \left(\frac{x}{2}ight) \cosh \left(\frac{x}{2}ight)}{2 \$$cosh^2$$ \left(\frac{x}{2}ight)} \, dx$$, which simplifies to $$\int \frac{\sinh \left(\frac{x}{2}ight)}{\cosh \left(\frac{x}{2}ight)} \, dx. $$Recognize that $$\frac{\sinh u}{\cosh u} = 	anh u$$, where $$u = \frac{x}{2}. $$Use substitution $$u = \frac{x}{2}$$, so $$dx = 2 du$$, and rewrite the integral as $$\int 	anh u \cdot 2 \, du = 2 \int 	anh u \, du. $$Then, integrate $$	anh u$$ using its known antiderivative.

37–56. Integrals Evaluate each integral.
∫ sinh x / (1 + cosh x) dx

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Key Concepts

Hyperbolic Functions

Integration Techniques for Rational Functions

Substitution Method