ln-x-sin-1-x-dx-

Question Number 114094 by Lordose last updated on 17/Sep/20

\int \ln (x) \sin^{- 1} (x) dx

Answered by Olaf last updated on 17/Sep/20

u^{'} = \ln x, u = x \ln x - x

v = \arcsin x, v^{'} = \frac{1}{\sqrt{1 - x^{2}}}

(x \ln x - x) \arcsin x - \int \frac{x \ln x - x}{\sqrt{1 - x^{2}}} d x

x \ln (\frac{x}{e}) \arcsin x + \int \ln (\frac{x}{e}) \frac{- 2 x}{2 \sqrt{1 - x^{2}}} d x

u = \ln (\frac{x}{e}), u^{'} = \frac{1}{x}

v^{'} = \frac{- 2 x}{2 \sqrt{1 - x^{2}}}, v = \sqrt{1 - x^{2}}

x \ln (\frac{x}{e}) \arcsin x + \ln (\frac{x}{e}) \sqrt{1 - x^{2}} - \int \frac{1}{x} \sqrt{1 - x^{2}} d x

\ln (\frac{x}{e}) (x \arcsin x + \sqrt{1 - x^{2}}) - \int \frac{1}{x} \sqrt{1 - x^{2}} d x

x = sinu

\int \frac{1}{x} \sqrt{1 - x^{2}} d x =

\int \frac{1}{\sin u} \sqrt{1 - \sin^{2} u} . \cos u d u =

\int ε \frac{\cos^{2} u}{\sin u} d u = ε \int \frac{1 - \sin^{2} u}{\sin u} d u (ε = \pm 1)

= ε \int (\csc u - \sin u) d u =

ε (\frac{1}{2} \ln ∣ \frac{1 - \cos u}{1 + \cos u} ∣ + \cos u) =

ε (\frac{1}{2} \ln ∣ \frac{1 - \sqrt{1 - x^{2}}}{1 + \sqrt{1 - x^{2}}} ∣ + \sqrt{1 - x^{2}}) \dots

Commented by Lordose last updated on 17/Sep/20

Complete it sir

Answered by 1549442205PVT last updated on 17/Sep/20

Integrating by part we have

I = \int \ln (x) \sin^{- 1} (x) dx = \int (lnx - 1) \sin^{- 1} (x) dx + \int \sin^{- 1} (x) dx

= (xlnx - x) \sin^{- 1} (x) - \int x {[(lnx - 1) \sin^{- 1} (x)]}^{'} dx + \int \sin^{- 1} xdx

= (xlnx - x) - \int \frac{xlnx}{\sqrt{1 - x^{2}}} dx + \int \frac{x}{\sqrt{1 - x^{2}}} dx (1)

Since \int \frac{x}{\sqrt{1 - x^{2}}} dx = \sqrt{1 - x^{2}},

We just need must find \int \frac{xlnx}{\sqrt{1 - x^{2}}} dx

Put x = \sin θ \Rightarrow dx = \cos θ d θ

\int \frac{xlnx}{\sqrt{1 - x^{2}}} dx = \int \sin θ lnsin θ d θ = - \cos θ lnsin θ

- \int (- \cos θ) {(lnsin θ)}^{'} d θ = - \cos θ lnsin θ + \int \frac{\cos^{2} θ}{\sin θ} d θ

= - \cos θ lnsin θ + \int \frac{1 - \sin^{2} θ}{\sin θ} d θ =

= - \cos θ lnsin θ + \int (cosec θ - \sin θ) d θ

= - \cos θ lnsin θ + \ln (cosec θ - \cos θ) + \cos θ

= - \sqrt{1 - x^{2}} lnx + \ln (\frac{1 - \sqrt{1 - x^{2}}}{x}) + \sqrt{1 - x^{2}}

From (1) we get

I = (x l nx - x) \sin^{- 1} x + \sqrt{1 - x^{2}} lnx - \ln (\frac{1 - \sqrt{1 - x^{2}}}{x}) + C

Answered by MJS_new last updated on 17/Sep/20

again just because something went wrong above \dots

\int \ln x \arcsin x d x =

[by parts]

= (\sqrt{1 - x^{2}} + x \arcsin x) \ln x - \int \frac{\sqrt{1 - x^{2}}}{x} + \arcsin x d x

\int \frac{\sqrt{1 - x^{2}}}{x} d x = \sqrt{1 - x^{2}} + \ln \frac{x}{1 + \sqrt{1 - x^{2}}}

\int \arcsin x d x = \sqrt{1 - x^{2}} + x \arcsin x

\Rightarrow

\int \ln x \arcsin x d x =

= \ln \frac{1 + \sqrt{1 - x^{2}}}{x} + (\sqrt{1 - x^{2}} + x \arcsin x) \ln x - x \arcsin x - 2 \sqrt{1 - x^{2}} + C

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