Question

Evaluate integration of cosh²xdx​

Answer 1

We're asked to evaluate,

$\displaystyle\sf{\longrightarrow I=\int\cos (h^2x)\ dx\quad\quad\dots (1)}$

Let $\displaystyle\sf {h}$ be constant wrt $\displaystyle\sf {x.}$

Let,

$\displaystyle\sf{\longrightarrow u=h^2x}$

$\displaystyle\sf{\longrightarrow du=h^2\,dx}$

$\displaystyle\sf{\longrightarrow dx=\dfrac {1}{h^2}\,du}$

Then (1) becomes,

$\displaystyle\sf{\longrightarrow I=\dfrac {1}{h^2}\int\cos u\ du}$

We know $\displaystyle\sf {\int\cos x\ dx=\sin x.}$

$\displaystyle\sf{\longrightarrow I=\dfrac {1}{h^2}\sin u+c}$

$\displaystyle\sf{\longrightarrow \underline {\underline {I=\dfrac {1}{h^2}\sin (h^2x)+c}}}$

where $\displaystyle\sf {c}$ is an arbitrary constant.

Answer 2

To Find:

$\\→I= \int \cosh ^{2} (x) dx$

Using the defined value of the hyperbolic cosine function.

i.e.,

$→\cosh(x) = \frac{ {e}^{x} + {e}^{ - x} }{2}$

$→{ \cosh}^{2} (x) = \frac{ {e}^{2x} + 2 + {e}^{ - 2x} }{4}$

$\implies {cosh}^{2} (x) = \frac{1 + \cosh(2x)}{2}$

Multiplying both sides by dx and then integrating both sides.

$\\ →I = \frac{1}{2} (\int dx + \int \cosh(2x)dx)$

$\\→ <u>I = \frac{2x + \sinh(2x) }{4} + C \: ; \: \: C ∈ ℝ</u>$