 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/17/a/ii/solution

sinh^{2} x = cosh^{2} x - 1

. With

u = cosh x

and

d u = sinh x d x

\int \frac{s i n h ^{3} x}{c o s h ^{2} x} d x = \int \frac{u ^{2} - 1}{u ^{2}} d u = \int (1 - u^{- 2}) d u = cosh x + \frac{1}{cosh x} + C .

For real

x

, the hyperbolic cosine never vanishes. Differentiating the result gives

sinh x (1 - sech^{2} x) = sinh^{3} x / cosh^{2} x

, checking the sign of the reciprocal term.

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