Answer to Question #247647 in Calculus for JaytheCreator

Question #247647

Find (i) ∫ ln π‘₯ 𝑑π‘₯, (ii) ∫ 𝑒 βˆ’π‘₯ cos 2π‘₯ 𝑑π‘₯. Hint: Let 𝐼 = ∫ 𝑒 π‘₯ cos 2π‘₯ 𝑑π‘₯ and use parts twice, (iii) sinβˆ’1 π‘₯ 𝑑π‘₯.


1
Expert's answer
2021-10-19T10:43:30-0400

Solution:

(i)

"I=\\int\\ln x\\ dx\n\\\\=\\int1\\times\\ln x\\ dx"

Using integration by parts,

"I=\\ln x\\int1\\ dx-\\int(\\dfrac d{dx}\\ln x\\int1dx)dx\n\\\\=\\ln x \\times x-\\int(\\dfrac 1{x}\\times x)dx\n\\\\=x\\ln x -x+C"

(ii)

"I=\\int e^{-x}\\cos 2x\\ dx" ...(A)

Using integration by parts,

"I=\\cos 2x\\int e^{-x} dx-\\int(\\dfrac d{dx}\\cos 2x \\int e^{-x}dx)dx\n\\\\=\\cos 2x \\times- e^{-x} -\\int(-2\\sin 2x \\times-e^{-x})dx\n\\\\=- e^{-x}\\cos 2x -2\\int(\\sin 2x \\times e^{-x})dx"

Again, using integration by parts,

"I=- e^{-x}\\cos 2x -2[\\sin 2x \\int e^{-x}dx-\\int(\\dfrac d{dx}\\sin 2x \\times \\int e^{-x}dx)dx]"

"=- e^{-x}\\cos 2x -2[-e^{-x}\\sin 2x -\\int(2\\cos 2x \\times - e^{-x})dx]\n\\\\=- e^{-x}\\cos 2x +2e^{-x}\\sin 2x -4\\int(\\cos 2x \\times e^{-x})dx"

"I=- e^{-x}\\cos 2x +2e^{-x}\\sin 2x -4I+C" [Using (A)]

"5I=- e^{-x}\\cos 2x +2e^{-x}\\sin 2x +C\n\\\\\\Rightarrow I=\\dfrac15(2e^{-x}\\sin 2x- e^{-x}\\cos 2x +C)"

(iii)

"I=\\int \\sin^{-1}x \\ dx\n\\\\=\\int 1\\times \\sin^{-1}x \\ dx"

Using integration by parts,

"I=\\sin^{-1}x\\times\\int 1 \\ dx-\\int(\\dfrac d{dx}\\sin^{-1}x\\int1 dx)dx\n\\\\=\\sin^{-1}x\\times x-\\int(\\dfrac 1{\\sqrt{1-x^2}}\\times x)dx"

"=x\\sin^{-1}x-\\dfrac12\\int\\dfrac {2x}{\\sqrt{1-x^2}}dx\n\\\\=x\\sin^{-1}x-\\dfrac12I_1+C"

Now, "I_1=\\int\\dfrac {2x}{\\sqrt{1-x^2}}dx"

Put "1-x^2=t"

"\\Rightarrow 2xdx=-dt"

So, we get,

"I_1=-\\int\\dfrac {1}{\\sqrt{t}}dt=-\\int t^{-1\/2} dt=-\\dfrac{t^{1\/2}}{1\/2}=-2\\sqrt t=-2\\sqrt{1-x^2}"

Thus, "I=x\\sin^{-1}x-\\dfrac12(-2\\sqrt{1-x^2})+C"

"\\Rightarrow I=x\\sin^{-1}x+\\sqrt{1-x^2}+C"


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