Evaluate the integral. ∫sec⁡2xcsc⁡2xdx

Question

Evaluate the integral.  ∫sec⁡2xcsc⁡2xdx

Corgnatiui · Accepted Answer

Step 1
Solution:
Consider the given integral:

I = \int \frac{\sec 2 x}{\csc 2 x} d x

Now, rewrite the integral:

I = \int \frac{1}{\cos 2 x} \frac{\sin 2 x}{1} d x

I = \int \frac{\sin 2 x}{\cos 2 x} d x

Let

u = \cos 2 x

\Rightarrow d u = - 2 \sin 2 x d x

\Rightarrow \sin 2 x d x = \frac{d u}{- 2}

Step 2
Therefore,

I = \int \frac{d u}{- 2 u}

I = - \frac{1}{2} \int \frac{d u}{u} = - \frac{1}{2} \ln | u |

I = - \frac{1}{2} \ln | \cos 2 x | + C

This is the required result, where C is the constant of integration.

Esther Phillips · Accepted Answer

∫sec⁡(2x)csc⁡(2x)dx  =12∫sec⁡(u)csc⁡(u)du  ∫sec⁡(u)csc⁡(u)du  =∫sin⁡(u)cos⁡(u)du  =−∫1vdv  ∫1vdv  =ln⁡(v)  −∫1vdv  =−ln⁡(v)  =−ln⁡(cos⁡(u))  12∫sec⁡(u)csc⁡(u)du  =−ln⁡(cos⁡(u))2  =−ln⁡(cos⁡(2x))2  ∫sec⁡(2x)csc⁡(2x)dx  =−ln⁡(|cos⁡(2x)|)2+C

karton · Accepted Answer

∫sec⁡(2x)csc⁡(2x)dx ∫1cos⁡(2x)1sin⁡(2x)dx ∫sin⁡(2x)cos⁡(2x)dx ∫−12tdt −12×∫1tdt 12×ln⁡(|t|) −12×ln⁡(|cos⁡(2x)|) Add C Answer: −12×ln⁡(|cos⁡(2x)|)+C

Evaluate the integral. \int \frac{\sec 2x}{\csc 2x}dx

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Answer & Explanation

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