Find the indefinite integral. ∫csc⁡2xdx

Question

Find the indefinite integral.  ∫csc⁡2xdx

eskalopit · Accepted Answer

Step 1 Consider the given integral. I=∫csc⁡2xdx Replace u=2x in the integral. u=2x du=2dx 12du=dx Step 2 Put the value in the integral. I=12∫csc⁡u(csc⁡u+cot⁡u)csc⁡u+cot⁡udu =12∫(csc2u+csc⁡u⋅cot⁡u)csc⁡u+cot⁡udu Putcsc⁡u+cot⁡u=t in the integral. csc⁡u+cot⁡u=t −(csc2u+cot⁡u⋅csc⁡u)du=dt (csc2u+cot⁡u⋅csc⁡u)du=−dt Step 3 Substitute all the values in the integral. I=−12∫dtt =−12ln|t|+c =−12ln|csc⁡u+cot⁡u|+c =−12ln|1+cos⁡usin⁡u|+c =−12ln|1+cos⁡2xsin⁡2x|+c =−12ln|2cos2x2sin⁡xcos⁡x|+c =12ln|tan⁡x|+c

Jillian Edgerton · Accepted Answer

∫csc⁡(2x)dx  Substitution u=2x⇒dudx=2⇒dx=12du:  =12∫csc⁡(u)du  Now we calculate:  ∫csc⁡(u)du  We expand the fraction by csc⁡(u)+cot⁡(u):  =∫csc⁡(u)(csc⁡(u)+cot⁡(u))csc⁡(u)+cot⁡(u)du  We use the distributive property:  =∫csc2(u)+cot⁡(u)csc⁡(u)csc⁡(u)+cot⁡(u)du  Substitution v=csc⁡(u)+cot⁡(u)⇒dvdu=−csc2(u)−cot⁡(u)csc⁡(u)⇒  du=1−csc2(u)−cot⁡(u)csc⁡(u)dv:  =−∫1vdv  Now we calculate:  ∫1vdv  This is the well-known tabular integral:  =ln⁡(v)  We substitute the already calculated integrals:  −∫1vdv  =−ln⁡(v)  Reverse replacement v=csc⁡(u)+cot⁡(u):  =−ln⁡(csc⁡(u)+cot⁡(u))  We substitute the already calculated integrals:  12∫csc⁡(u)du  =−ln⁡(csc⁡(u)+cot⁡(u))2  Reverse replacement u=2x:  =−ln⁡(csc⁡(2x)+cot⁡(2x))2  The problem has been solved. Applying the modulus to the logarithm argument expands its range:  ∫

nick1337 · Accepted Answer

Find the indefinite integral. \int \csc 2xdx

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