# Evaluating an integral using a trig substitution

Problem:
Evaluate the following integral: $$\int \frac{\sqrt{x}}{1+x} \, dx$$ Answer:
Let $$I$$ be the integral we are evaluating. We are going to use the substitution $$\tan u = \sqrt{x}$$. We have: \begin{align*} \sec^2{u} \,\, du &= \left( \frac{1}{2} \right) x^{ - \frac{1}{2} } \, dx \\ I &= 2 \int \frac{ \left( \frac{1}{2} \left( \frac{x}{ \sqrt{x} }\right) \right) }{1+x} \, dx \\ I &= 2 \int \frac{\tan^2{u}} {\tan^2{u}+ 1} \, du = 2 \int \frac{ \tan^2 u } { \sec^2 u } \, du \\ I &= 2 \int \sin^2 u \, du \\ \end{align*} Now recall the trig idenity: $$\sin^2 \theta = \frac{2 - \cos{2 \theta} }{2}$$ \begin{align*} I &= 2 \int \frac{2 - \cos{2 u} }{2} \, du = \int 2 - \cos{2 u} \, du \\ I &= 2u - \frac{\sin{2u}}{2} + C = 2u - \frac{\sqrt{ 1 - \cos^2{2u}}}{2} + C \end{align*} The book's answer is: $$\int \frac{\sqrt{x}}{1+x} \, dx = 2\sqrt{x} - 2\tan{( \sqrt{x}) } + C$$ I made a mistake in entering the book's answer. Here is what it should be:
The book's answer is: $$\int \frac{\sqrt{x}}{1+x} \, dx = 2\sqrt{x} - 2\tan^{-1}{( \sqrt{x}) } + C$$ Therefore, I conclude my answer is wrong. Where did I go wrong?

• You have to use a trig substitution ? $u=\sqrt x$ is a better substitution Mar 8, 2020 at 2:22
• I do not have to use a trig substitution but when I tired $u = \sqrt{x}$ it did not work out so well. I am going to try again.
– Bob
Mar 8, 2020 at 2:25
• What do you mean it didnt work ? $u=\sqrt x \implies du =\frac 12 \frac {dx}{\sqrt x}$ Mar 8, 2020 at 2:27
• The book's answer is WRONG! "tan" should be "arctan" Mar 8, 2020 at 2:28
• The book has the right answer. I typed it in wrong. I will fix the post.
– Bob
Mar 8, 2020 at 2:32

$$I=\int \frac{\sqrt{x}}{1+x} \, dx$$ Substitute $$u=\sqrt x \implies dx=2udu$$ $$I=2\int \frac{{u^2}}{1+u^2} \, du=2\int \frac{{u^2+1-1}}{1+u^2} \, du$$ $$I =2 \left(u-\int \frac{du}{1+u^2} \right)$$ $$I =2(u-\arctan u) +C =2(\sqrt{x}-\arctan \sqrt{x}) +C$$

• What I missed was the trick of writing $u^2$ as $u^2 + 1 - 1$.
– Bob
Mar 8, 2020 at 2:44
• @bob i see. it's a well known trick. This substitution is far simple than a trig substitution. You make less mistakes Mar 8, 2020 at 2:45

You made a mistake during the substitution. With $$\tan u = \sqrt{x}$$, you get $$x = \tan^2u$$ and $$dx = 2\tan u \sec^2udu$$. Then,

$$\int \frac{\sqrt{x}}{1+x}dx = \int \frac{\tan^2u \cdot 2\sec^2u}{1+\tan^2u}du = 2\int \tan^2udu = 2\int (\sec^2u-1)du$$

$$=2(\tan u-u)+C = 2(\sqrt x - \tan^{-1}\sqrt x)+C$$

\begin{aligned} & \int \frac{\sqrt{x}}{1+x} d x \\ =& \int \frac{x}{(1+x) \sqrt{x}} d x \\ =& 2 \int \frac{x}{1+x} d(\sqrt{x}) \\ =& 2\int\frac{1+x-1}{1+x} d (\sqrt{x})\\ =&\left.2 \int 1 d (\sqrt{x})-\int \frac{d (\sqrt{x})}{(\sqrt{x})^{2}+1}\right] \\ =& 2(\sqrt{x}-\arctan \sqrt{x})+C \end{aligned}