Question

Find (i) ∫ ln 𝑥 𝑑𝑥, (ii) ∫ 𝑒 −𝑥 cos 2𝑥 𝑑𝑥. Hint: Let 𝐼 = ∫ 𝑒 𝑥 cos 2𝑥 𝑑𝑥 and use parts twice, (iii) sin−1 𝑥 𝑑𝑥.

Answer 1

Solution:

(i)

$I=\int\ln x\ dx \\=\int1\times\ln x\ dx$

Using integration by parts,

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

(ii)

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

Using integration by parts,

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

Again, using integration by parts,

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

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

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

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

(iii)

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

Using integration by parts,

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

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

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

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