Answer to Question #184744 in Calculus for Angelo

Question

Evaluate the the Integration by Trigonomentric Substitution:

1. ∫ √x² -1 dx/x²

2. ∫ dx/√6x+x²

3. ∫ dx/x√x⁴ -4

4. ∫ x² dx/√9-4x²

Answer 1

$1) \int \frac{\sqrt {x^2-1}}{x^2}dx=I\\ substitute\\ x=\frac{1}{\sin t}\\ dx=-\frac{\cos t dt}{\sin^2t}\\ I=\int\frac{\sqrt{\frac{1}{\sin^2 t}-1}}{\frac{1}{\sin t}} \frac{-\cos t dt}{\sin^2t}=\\ =\int\sqrt{\frac{1-\sin^2t}{\sin^2t}}(-\cos t)dt=\\ =-\int\frac{\cos^2t}{\sin t }dt=\\ =-\int\frac{1-\sin^2t}{\sin t}dt=\\ =\int(\sin t- \frac{1}{\sin t}) dt=\\ =-\cos t -\ln|\tan\frac{t}{2}| + c=\\ =- \cos(\arcsin \frac{1}{x}) - \ln|\tan \frac{1}{2}\arcsin \frac{1}{x}| +c\\ 2) \int\frac{dx}{\sqrt{6x+x^2}}=I\\ substitute\\ \sqrt{6x+x^2}=t+x\\ x=\frac{t^2}{6-2t}\\ dx=\frac{12t-2t^2}{(6-2t)^2}dt\\ x+t=\frac{-t^2+6t}{6-2t}\\ then\\ I=\int\frac{\frac{12t-2t^2}{(6-2t)^2}dt}{\frac{-t^2+6t}{6-2t}}==\\ =\int \frac{2 dt}{6-2t}=-\ln|6-2t|+c=\\ =-\ln|6-2(\sqrt{6x+x^2}-x)|+c\\ 3) \int\frac{dx}{x\sqrt{x^4-4}}=I\\ substitute\\ x^2=t\\ t=\frac{2}{\sin z}\\ dt=-\frac{\cos z dz}{\sin^2z}\\ I=\int\frac{xdx}{x^2 \sqrt{x^4-4}}=\\ =\frac{1}{2}\int\frac{dx^2}{x^2\sqrt{x^4-4}}=\\ =\frac{1}{2}\int\frac{dt}{t\sqrt{t^2-4}}=\\ =\frac{1}{2}\int\frac{-\frac{\cos z dz}{\sin^2z}dz}{\frac{2}{\sin z}\sqrt{\frac{4}{\sin^2z}-4}}=\\ =-\frac{1}{4}\int\frac{\frac{\cos z}{\sin z}dz}{\frac{2\cos z}{\sin z}}=\\ =-\frac{1}{8}\int dz=-\frac{1}{8}z+c=-\frac{1}{8}\arcsin \frac{2}{x^2}+c\\ 4) \int\frac{x^2dx}{\sqrt{9-4x^2}}=I\\ substitute\\ x=\frac{3}{2}\sin t\\ dx=\frac{3}{2}\cos t dt\\ I=\int\frac{\frac{9}{4}\sin^2t \cdot\frac{3}{2}\cos t dt}{\sqrt{9-4 \cdot \frac{9}{4} \sin^2t }}=\\ =\int\frac{\frac{27}{8}\sin^2t \cos t dt}{3 \cos t}=\\ =\frac{9}{8}\int\sin^2t dt=\\ =\frac{9}{16}\int(1-\cos 2t) dt=\\ =\frac{9}{16}(t-\frac{1}{2} \sin2t) + c=\\ =\frac{9}{16}(\arcsin\frac{2x}{3}-\frac{1}{2} \sin(2\arcsin\frac{2x}{3})) + c$