Question

Evaluate integration of cosh²xdx​

Answer 1

Answer Reffered To the Attachment

Answer 2

Step-by-step explanation:

We have to evaluate :

$\sf→ \: l \: ∫ \cos(h {}^{2} x) dx </strong><strong> </strong><strong>$

Let h be constant wrt x.

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Let,

$\sf → \: u = {h}^{2} x \\ \sf → \: du = h {}^{2} dx \\ \sf → \: dx = \frac{1}{ {h}^{2} } du$

Now 1 becomes ,

$\sf → \: l = \frac{1}{ {h}^{2} } ∫ \: \cos \: u \: du$

As we know

$\sf →\: ∫ \: \cos \: x \: dx = \sin \: x$

$\sf→ \: l = \frac{1}{h ^{2}} \sin \: u + c$

$\underline{ \sf → \: l = \frac{1}{ {h}^{2} } \sin(h {}^{2} x) + c}$

Where c is an arbitrary constant.

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