My textbook asked me to evaluate the integral $\int_1^5\left(\frac{x}{\sqrt{2x-1}}\right)$ using u-substitution. I rewrote the integrand as $x*(2x-1)^{-\frac{1}{2}}$, but I soon became stuck because I couldn't settle on what was $g'(x), g(x),$ or $F(x)$ (I didn't know how to define u or du). I checked the book (this was an example problem) and it said to set u = $\sqrt{2x-1}$. How did it come to this conclusion? Thanks for your help.
Question about evaluating a Definite Integral w/ U-Substitution
calculusintegration
Related Solutions
There was a typo in the current post. After enforcing the substitution $2x=3\tan \theta$, the integral ought to read
$$\begin{align}I&=\int_{\arctan(2/3)}^{\arctan(4/3)}\frac{\frac32 \sec^2\theta}{\frac94 \tan^2\theta\sqrt{9\tan^2\theta+9}}d\theta\\\\ &=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\frac{ \sec^2\theta}{ \tan^2\theta\,\sec \theta}d\theta\\\\ &=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\frac{ \sec^2\theta}{ \tan^2\theta\,\sec \theta}d\theta\\\\ &=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\cot \theta \csc \theta d\theta\\\\ &=\frac29 \left.(-\csc \theta)\right|_{\arctan(2/3)}^{\arctan(4/3)}\\\\ &=\frac{\sqrt{13}}{9}-\frac{5}{18}\end{align}$$
NOTES:
Remark 1: When making a substitution of variables in a definite integral, the limits of integration change accordingly. In this example, the substitution was $x=\frac32 \tan \theta$. When $x=1$ at the lower limit, $\tan \theta =\frac23\implies \theta =\arctan(2/3)$. Similarly, when $x=2$ at the upper limit, $\tan \theta =\frac43\implies \theta =\arctan(4/3)$.
Remark 2:
To evaluate $\sin (\arctan(2/3))$, we recall that the arctangent is an angle whose tangent is $2/3$. A picture sometimes facilitates the analysis wherein we draw a right triangle with vertical side of length $2$ and horizontal side of length $3$ forming a right angle.
Note that the angle the hypotenuse makes with the horizontal side is $\arctan(2/3)$. Inasmcuh as the hypotenuse is of length $\sqrt{2^2+3^2}=\sqrt{13}$, we see $\sin(\arctan(2/3))=\frac{2}{\sqrt{13}}$ and thus $\csc (\arctan(2/3))=\frac{\sqrt{13}}{2}$.
Definite Integral – How to Solve the Definite Integral $\\int_{0}^{a}\\frac{x^4dx}{\\sqrt{a^2-x^2}}$
With $x=a\sin t$, which isn't quite what you said you tried, the integral is$$\begin{align}a^4\int_0^{\pi/2}\sin^4tdt&=\frac14a^4\int_0^{\pi/2}(1-\cos2t)^2dt\\&=\frac14a^4\int_0^{\pi/2}(1-2\cos2t+\cos^22t)dt\\&=\frac18a^4\int_0^{\pi/2}(3-4\cos2t+\cos4t)dt\\&=\frac18a^4[3t-2\sin2t+\tfrac14\sin4t]_0^{\pi/2}\\&=\frac{3\pi}{16}a^4.\end{align}$$
Best Answer
They are trying to get you to use the trick that setting $$u=\sqrt{2x-1}\implies\frac{u^2+1}{2}=x\implies u \space du=dx$$ Then $$I=\int_{x=1}^{x=5}\frac{x}{\sqrt{2x-1}}dx=\int_{u=1}^{u=3}\frac{\left(\frac{u^2+1}{2}\right)}{u}\cdot u\space du=\frac{1}{2}\int_1^3(u^2+1)\space du$$