How to evaluate the integral: $\int\ln x\;\sin^{-1} x\, \operatorname d\!x$?

Question

Can the following integral integrated by parts?
$$\int\ln x\;\sin^{-1} x\, \operatorname d\!x$$

Answer 1

Integration by parts formula

$$\int u(x)\underset{dv}{\underbrace{v^{\prime }(x)dx}}=u(x)v(x)-\int v(x) \underset{du}{\underbrace{u^{\prime }(x)dx}}.$$

If we apply the LIATE rule to

$$\int \ln x\arcsin x\,dx,$$

we should choose

$$u =\ln x,\quad dv=\arcsin x\,dx.$$

Normally there is a choice of the terms $u,v'$ of the integrand $uv'$ that makes integration much easier. However in the present case the other possible choice $u=\arcsin x,v'=\ln x$ leads to integrals of similar difficulties.

$$\begin{eqnarray*} I &=&\int \ln x\arcsin x\,dx, \qquad \text{(}v =\int \arcsin x\,dx\quad \text{see evaluation below)} \\ &=&\left( \ln x\right) \left( x\arcsin x+\sqrt{1-x^{2}}\right) -\int \frac{ 1}{x}\left( x\arcsin x+\sqrt{1-x^{2}}\right) \,dx \\ &=&\left( \ln x\right) \left( x\arcsin x+\sqrt{1-x^{2}}\right) -\int \arcsin x\,dx-\int \frac{1}{x}\sqrt{1-x^{2}}\,dx \\ &&\text{(see evaluation of the last integral below)} \\ &=&\left( \ln x\right) \left( x\arcsin x+\sqrt{1-x^{2}}\right) -\left( \sqrt{1-x^{2}}+x\arcsin x\right) \\ &&-\sqrt{1-x^{2}}-\frac{1}{2}\ln \frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}+1}+C. \\ &=&\left( -1+\ln x\right) \left( x\arcsin x+\sqrt{1-x^{2}}\right) -\sqrt{ 1-x^{2}} -\frac{1}{2}\ln \frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}+1}+C. \end{eqnarray*}$$

• Evaluation of $\int \arcsin x\,dx$ by substitution $$\begin{eqnarray*} \int \arcsin x\,dx &=&\int u\cos u\,du,\quad u=\arcsin x \\ &=&\cos u+u\sin u \\ &=&\sqrt{1-x^{2}}+x\arcsin x+C. \end{eqnarray*}$$
• Evaluation of $\int \frac{1}{x}\sqrt{1-x^{2}}\,dx$ by substitution and partial fractions $$\begin{eqnarray*} \int \frac{1}{x}\sqrt{1-x^{2}}\,dx &=&-\int \frac{u^{2}}{1-u^{2}}\,du,\qquad u=\sqrt{1-x^{2}} \\ &=&u+\frac{1}{2}\ln \left( u-1\right) -\frac{1}{2}\ln \left( u+1\right) \\ &=&u+\frac{1}{2}\ln \frac{u-1}{u+1} \\ &=&\sqrt{1-x^{2}}+\frac{1}{2}\ln \frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}+1}+C. \end{eqnarray*}$$

Answer 2

Following J.M.'s hint, you may want to prove, again integrating by parts, that $$\int\log x\,dx=x\log x-x+C\,\,,\,\,C=\,\,\text{a constant}$$and perhaps also to note that $$\int \frac{x}{\sqrt{1-x^2}}dx=-\frac{1}{2}\int\frac{-2x}{\sqrt{1-x^2}}dx$$

Answer 3

Let I = ∫ ln x * sin−1 x dx,

= x(ln x - 1) * arcsin x - ∫ x(ln x - 1) * 1/√(1 – x^2) dx

= x(ln x - 1) * arcsin x - ∫(ln x - 1) * x / √(1 – x^2) dx

= x(ln x - 1) * arcsin x + (ln x - 1) * √(1 – x^2) – ∫ (1/x) * √(1 – x^2) dx

Call this last integral I1.

Put u = x^2; then du/dx = 2x

∴ I1 = ∫ (1/(2u)) * √(1 – u) * du/dx dx

Put u = 1 – v^2, v ≥ 0; so du/dv = -2v

∴ I1 = ½ ∫ 1/(1 – v^2) * (-2v^2) dv

v^2 / (v^2 – 1) = (v^2 – 1 + 1) / (v^2 – 1) = 1 + 1 / (v^2 – 1)

= 1 + ½ (1/(v – 1) - 1/(v + 1))

So I1 = v + ½ ln (A * (v – 1) / (v + 1)), where A is the arbitrary constant

= √(1 – u) + ½ ln (A * (√(1 – u) – 1) / (√(1 – u) + 1)),

= √(1 – x^2) + ½ ln (A * (√(1 – x^2) – 1) / (√(1 – x^2) + 1)),

Add this to the other parts, and you are done.

I at first thought it could not be done in elementary functions; but there we are!)