# integration with trigonometric substitution, is my result correct?

$$\int\frac{\ln x}{x\sqrt{1-4\ln x-\ln^{2}x}}\ dx\left | u=\ln x,\ du=\frac{1}{x},\ x du=dx \right | \\ \int \frac{u}{x\sqrt{-u^{2}-4u+1}}\ xdu\ = \int \frac{u}{\sqrt{-u^{2}-4u+1}}\ du \\ \left | -u^{2}-4u+1=-(u^{2}+4u-1) \right | \\ \\ \left | -(u+2)^{2}+5=-u^{2}-4u+1 \right | \\ \\\int \frac{u}{\sqrt{-(u+2)^{2}+5}}\ du\left | \sqrt{-x^{2}+a^{2}}=\sqrt{a^{2}-x^{2}} \implies a\sin\theta \right | \\ u+2=x\implies\sqrt{5}\sin\theta,\\ dx=\sqrt{5}\cos\theta \ d\theta \\ \int \frac{u+2-2}{\sqrt{-(\sqrt{5}\sin\theta)^{2}+5}}\cdot\sqrt{5}\cos\theta \ d\theta \\ \int \frac{\sqrt{5}\sin\theta-2}{\sqrt{-5\sin^{2}\theta+5}}\cdot\sqrt{5}\cos\theta \ d\theta \\ \int \frac{\sqrt{5}\sin\theta-2}{\sqrt{5(1-\sin^{2}\theta)}}\cdot\sqrt{5}\cos\theta \ d\theta \\ \int \frac{\sqrt{5}\sin\theta-2}{\sqrt{5}\sqrt{\cos^{2}\theta}}\cdot\sqrt{5}\cos\theta \ d\theta \\ \int \frac{\sqrt{5}\sin\theta-2}{\sqrt{5}\cdot \cos\theta}\cdot\sqrt{5}\cos\theta \ d\theta \\ \int \sqrt{5}\sin\theta-2 \ d\theta =-\sqrt{5}\cos\theta-2\theta+c \\ \sqrt{5}\sin\theta=x, \ \sin\theta=\frac{x}{\sqrt{5}}, \ x=u+2 \implies u=\ln x, \ \theta =\sin^{-1}(\frac{\ln x+2}{\sqrt{5}})$$

• Welcome to MSE. I suggest that you type \sin, \cos and \ln in order to get $\sin$, $\cos$, and $\ln$ respectively. May 16, 2019 at 12:42

Yes it's correct. But avoid using $$x$$ twice for different meanings, as it may lead to confusion.

You have found $$\theta$$. Now, $$\cos\theta = \sqrt{1-\sin^2\theta}$$

$$\sqrt 5\cos\theta = \sqrt{5(1-\sin^2\theta)} = \sqrt{5\bigg(1 - (\frac{u+2}{\sqrt5}\bigg)^2}) = \sqrt{5-(u+2)^2} = \sqrt{-u^2-4u+1}$$

$$\sqrt 5\cos\theta = \sqrt{-\ln^2(x) - 4\ln(x) + 1}$$

Thus the final solution becomes,

$$I = -\sqrt{-\ln^2(x) - 4\ln(x) + 1} - 2sin^{-1}\bigg(\frac{\ln(x)+2}{\sqrt5}\bigg)+c$$

• $but\ if \ sin\theta =\frac{lnx+2}{\sqrt{5}}\ and\ if\ sin=\frac{a}{c}\\ c^{2}=a^{2}+b^{2} \\ \\ \sqrt{5}^{2}=(lnx+2)^{2}+b^{^{2}}\\ \\ b=\sqrt{5-(lnx+2)^{2}} \\ \\ cos=\frac{b}{c}=\frac{\sqrt{5-(lnx+2)^{2}} }{\sqrt(5)}\\ \\ -\sqrt{5}cos\theta -2\theta +c \\ \\ -\sqrt{5}\cdot \frac{\sqrt{5-(lnx+2)^{2}} }{\sqrt(5)}-2\cdot sin^{-1}(\frac{lnx+2}{\sqrt{5}})+c$ May 16, 2019 at 14:33
• Yes, after that $5 - (lnx+2)^2 =5 - ln^2(x) - 4ln(x) - 4 = -ln^2(x) - 4ln(x) +1$ . Also cancel $\sqrt 5$ from Nr. and Dr. May 16, 2019 at 14:40
• yeah I didnt do the operations til the end it matches, ok thanks May 16, 2019 at 14:43