Integrate $ \int \sinh 2x \cosh x dx$

Question

I want to integrate

$$\int \sinh 2x \cosh x dx$$

my steps was as follow:

$$ \int \sinh 2x \cosh x dx = \\ \sinh 2x . \sinh x - 2 \int \cosh 2x \sinh x dx = \\ \sinh 2x . \sinh x - 2 (\cosh 2x . \cosh x - 2 \int \sinh 2x \cosh x dx) \\\equiv -4 \int \sinh 2x \cosh x dx = \sinh 2x . \sinh x - 2 \cosh 2x . \cosh x \\\equiv \int \sinh 2x \cosh x dx = -\frac{1}{4}\sinh 2x . \sinh x + \frac{2}{4} \cosh 2x . \cosh x = \frac{\cosh^{3}}{2} $$

But the result shown here is $2/3 \cosh^3(x)$.

Wat am I doing wrong?

Answer 1

Before the penultimate line, you have

$$\int \sinh 2x\cosh x\,dx = \sinh 2x \sinh x - 2\cosh 2x \cosh x + 4\int \sinh 2x\sinh x\,dx.$$

You forgot the one on the left to obtain a factor $-4$ in the penultimate line where it should have been $-3$. Nothing serious, ordinary lapse of attention.

Answer 2

I always prefer to do a $u$-substitution rather than integration by parts. As an analogy to circular trig functions, $$\sin(2x) = 2\sin x \cos x$$ and it can be shown that $$\sinh(2x) = 2 \sinh x \cosh x$$ Using this identity, rewrite the integral to get: $$\int \sinh(2x) \cosh(x) \, dx = \int 2 \sinh(x)\cosh(x)\cosh(x) \, dx \\ = 2 \int \sinh(x) \cosh^2(x) \, dx$$ Now is the time for a $u$-substitution where we let $u = \cosh(x)$, and then $du = \sinh(x) \, dx$. Thus, $$2 \int \sinh(x) \cosh^2(x) \, dx = 2 \int u^2 \, du \\ = 2\left(\frac{u^3}{3} \right) + C$$ Back-substituting in the value of $u$ gives your final answer $$\frac{2}{3} \cosh^3(x) + C$$