# Integration with trigonometry substitution

How do you integrate:

$$\int \sqrt{\frac{(4x-3)}{1-x}}dx$$

hint given was $\frac{1}{1-x} = 4\sec^2(u)$ do i need to use trigonometric substitution for this? Even so, not sure how to solve it

After trying, the answer i got was $$u/2 +{1/2\sin u\cos u} + c$$ And after substitution to get back x, i got $$\frac{\arccos(2\sqrt(1-x))}{2} + \sqrt(1-x)\sqrt(4x-3) + c$$

Is this correct?

• Use $u$ substitution and write the square root as two square roots. So if $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ – Latin Wolf Mar 29 '18 at 13:48
• @ Latin Wolf, mind if you could elaborate more on using u substitution? – Isabella.T Mar 29 '18 at 14:02
• if you use the hint given in the problem you should end up with $\int \sin ^2udu$ – Lozenges Mar 29 '18 at 14:06
• @ Lozenges, how do i use the hint for the numerator? – Isabella.T Mar 29 '18 at 14:08
• from the hint $4-4x =\cos ^2u$ we get $4x-3=\sin ^2u$ and $4 \text{dx} = 2 \sin u \cos u \text{du}$ – Lozenges Mar 29 '18 at 14:11

I would Substitute $$t=\sqrt{\frac{4x-3}{x-1}}$$ then we get $$x=\frac{t^2-3}{t^2-4}$$ and $$dx=-\frac{2t}{(t^2-4)^2}dt$$ then we get $$-2\int \frac{t^2}{(t^2-4)^2}dt$$ Can you solve this? for your control we get $$-2(-1/4\, \left( t-2 \right) ^{-1}+1/8\,\ln \left( t-2 \right) -1/4\, \left( t+2 \right) ^{-1}-1/8\,\ln \left( t+2 \right) )+C$$
• @ Dr. Sonnhard Graubner if I am not wrong $\int \frac{t^2}{(t^2-4)^2} dt= \int \frac{1}{t^2-4} dt + 4\int \frac{1}{(t^2-4)^2} dt$ – Latin Wolf Mar 29 '18 at 13:58
• use that $$\frac{t^2}{(t^2-4)^2}=1/4\, \left( t-2 \right) ^{-2}-1/8\, \left( t+2 \right) ^{-1}+1/8\, \left( t-2 \right) ^{-1}+1/4\, \left( t+2 \right) ^{-2}$$ – Dr. Sonnhard Graubner Mar 29 '18 at 14:02
hint: $$\frac{4x-3}{1-x}=-\frac{4(x-1)+1}{x-1}=-4+\frac{1}{1-x}$$
Using the hint $$\frac{1}{1-x} = 4\sec^2(u)\implies x=\frac{1}{8} (7-\cos (2 u))\implies dx=\frac{1}{4} \sin (2 u)\,du$$ So, after simplifications, $$\int \sqrt{\frac{4x-3}{1-x}}\,dx=\int \sin (u) \cos (u) \sqrt{\tan ^2(u)}\,du$$ I am sure that you can take it from here.
• Isn't your last integrand equal to $\sin^2(u)$ – Latin Wolf Mar 30 '18 at 3:04