Question

using suitable substitution, evaluate
integrate from 1 to 4
x^3/square root (x^2+1 ) dx

the ans: 19.7125

Answer 1

\[\int_1^4\frac{x^3}{\sqrt{x^2+1}}\,dx=\int_1^4x\frac{x^2}{\sqrt{x^2+1}}\,dx\]
Let \(u=x^2+1\), then \(x^2=u-1\) and \(x\,dx=\dfrac{du}{2}\).
\[\frac{1}{2}\int_2^{17}\frac{u-1}{\sqrt{u}}\,du=\frac{1}{2}\int_2^{17}\left(u^{1/2}-u^{-1/2}\right)\,du\]

Answer 2

Another approach, using a trigonometric substitution. Let \(x=\tan t\), then \(dx=\sec^2t\,dt\).
\[\int_1^4\frac{x^3}{\sqrt{x^2+1}}\,dx=\int_{\arctan 1}^{\arctan4}\frac{\tan^3t\sec^2t}{\sqrt{\tan^2t+1}}\,dt=\int_{\arctan 1}^{\arctan4}\frac{(1-\cos^2t)\sin t}{\cos^4 t}\,dt\]
Substituting \(y=\cos t\) gives \(-dy=\sin t\,dt\):
\[\begin{align*}\int_{\arctan 1}^{\arctan4}\frac{(1-\cos^2t)\sin t}{\cos^4 t}\,dt&=-\int_{\cos(\arctan1)}^{\cos(\arctan4)}\frac{1-y^2}{y^4}\,dy\\\\&=\int^{\cos(\arctan4)}_{\cos(\arctan1)}\left(\frac{1}{y^2}-\frac{1}{y^4}\right)\,dy\\\\
&=\int^{1/\sqrt{17}}_{1/\sqrt2}\left(\frac{1}{y^2}-\frac{1}{y^4}\right)\,dy\end{align*}\]