How do you evaluate the integral #int arctansqrtx#?

Answer 1

#intarctan(sqrtx)dx=(x+1)arctan(sqrtx)-sqrtx+C#

#I=intarctan(sqrtx)dx#

Let #t=sqrtx#. Finding #dx# in a usable form is simpler if we first write that #t^2=x#, which then implies that #2tdt=dx#. This way we can plug in #dx# into the integral straight away. These substitutions yield:

#I=intarctan(t)(2tdt)=int2tarctan(t)dt#

Now use integration by parts. Let:

#{(u=arctan(t)" "=>" "du=1/(t^2+1)dt),(dv=2tdt" "=>" "v=t^2):}#

Then:

#I=t^2arctan(t)-intt^2/(t^2+1)dt#

Rewriting the integrand as #(t^2+1-1)/(t^2+1)=(t^2+1)/(t^2+1)-1/(t^2+1)=1-1/(t^2+1)# we see that

#I=t^2arctan(t)-(intdt-int1/(t^2+1)dt)#

Both of which are common integrals:

#I=t^2arctan(t)-t+arctan(t)+C#

Returning to #x# from #t=sqrtx#:

#I=xarctan(sqrtx)-sqrtx+arctan(sqrtx)+C#

#I=(x+1)arctan(sqrtx)-sqrtx+C#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Axel Alvarez

To evaluate the integral (\int \arctan(\sqrt{x}) , dx), we can use integration by parts. Let ( u = \arctan(\sqrt{x}) ) and ( dv = dx ). Then, ( du = \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}} , dx ) and ( v = x ).

Applying the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

we get:

[ \begin{aligned} \int \arctan(\sqrt{x}) , dx &= x\arctan(\sqrt{x}) - \int \frac{x}{1 + x} \cdot \frac{1}{2\sqrt{x}} , dx \ &= x\arctan(\sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{x}}{1 + x} , dx \end{aligned} ]

Now, for the remaining integral (\int \frac{\sqrt{x}}{1 + x} , dx), we can use a substitution method, letting ( t = \sqrt{x} ) which implies ( x = t^2 ) and ( dx = 2t , dt ). Substituting these into the integral:

[ \begin{aligned} \int \frac{\sqrt{x}}{1 + x} , dx &= \int \frac{t \cdot 2t}{1 + t^2} , dt \ &= 2 \int \frac{t^2}{1 + t^2} , dt \end{aligned} ]

This integral can be evaluated using a trigonometric substitution, letting ( t = \tan(\theta) ), so ( dt = \sec^2(\theta) , d\theta ). Substituting these, we have:

[ \begin{aligned} 2 \int \frac{t^2}{1 + t^2} , dt &= 2 \int \frac{\tan^2(\theta) \cdot \sec^2(\theta)}{1 + \tan^2(\theta)} , d\theta \ &= 2 \int \frac{\tan^2(\theta) \cdot \sec^2(\theta)}{\sec^2(\theta)} , d\theta \ &= 2 \int \tan^2(\theta) , d\theta \ &= 2 \int (\sec^2(\theta) - 1) , d\theta \ &= 2 (\tan(\theta) - \theta) + C \end{aligned} ]

Now, we can back-substitute and integrate to get the final result:

[ \begin{aligned} \int \arctan(\sqrt{x}) , dx &= x\arctan(\sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{x}}{1 + x} , dx \ &= x\arctan(\sqrt{x}) - \frac{1}{2} \left( 2(\tan(\theta) - \theta) \right) + C \ &= x\arctan(\sqrt{x}) - \tan(\theta) + \theta + C \end{aligned} ]

Remembering that ( x = t^2 ) and ( t = \tan(\theta) ), we can replace ( \tan(\theta) ) with ( \sqrt{x} ) and ( \theta ) with ( \arctan(\sqrt{x}) ):

[ \int \arctan(\sqrt{x}) , dx = x\arctan(\sqrt{x}) - \sqrt{x} + \arctan(\sqrt{x}) + C ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.