Turn on thread page Beta
 You are Here: Home >< Maths

is it possible to watch

1. integrate the square root of tan x?
if so how?
2. Make the substitution y2 = tan x

When you've done this, rearrange the numerator to be

2y2 = 2(1+y2) - 2

And go from there.
3. how do you go from there ?
4. y = √tanx

∫ (tanx)1/2.dx
y² = tanx
2y.dy/dx = sec²x
dy/dx = (sec²x)/2y
dy/dx = (sec²x)/(2√(tanx))
5. (Original post by Dekota)
y = √tanx

∫ (tanx)1/2.dx
y² = tanx
2y.dy/dx = sec²x
dy/dx = (sec²x)/2y
dy/dx = (sec²x)/(2√(tanx))
Elegant solution.
6. (Original post by Sumeet)
Elegant solution.
What solution?
7. It is not solution yet. It's not that easy. This will leave you at the integration below:
8. Since I had nothing better to do...

Let I = ∫ sqrt[tan(x)] dx. Then using Cexy's sub, we transform the integral into:
I = 2 ∫ u^2/(1+u^4)

Some tricks are in order.
I = 2 ∫ (u^2 + 1 - 1)/(1 + u^4) du
= 2 ∫ (u^2 + 1)/(u^4 + 1) du - 2 ∫ 1/(1 + u^4) du
= 2J - 2H

Where:
J = ∫ (u^2 + 1)/(u^4 + 1) du
= ∫ (1 + 1/u^2)/(u^2 + 1/u^2) du

Setting v=u-1/u with dv/dv=1+1/u^2, we get:
J = ∫ dv/(v^2 + 2) = (1/sqrt2) arctan(v/sqrt2) + C

And:
H = ∫ 1/(1 + u^4) du

Now note that (1 + u^4) = (1+u^2)^2 - 2u^2 = (1+u^2-sqrt(2)u)(1+u^2+sqrt(2)u).

So you can split the integrand in H using partial fractions. Eventually, and I mean eventually, you will be able to integrate H. And hence you can find ∫sqrt[tan(x)] dx.

Alternatively
I = 2 ∫ u^2/(1+u^4) du
could be split into something prettier using a dirty trick:
I = ∫ [(u^2 + 1) + (u^2 - 1)]/(1+u^4) du
= ∫ (u^2 + 1)/(u^4 + 1) du + ∫ (u^2 - 1)/(u^4 + 1) du
= ∫ (1 + 1/u^2)/(u^2 + 1/u^2) du + ∫ (1 - 1/u^2)/(u^2 + 1/u^2) du
= J + H*

We already did J, and H* can be done similarly if we use the substituion v=u+1/u instead.
9. (Original post by BCHL85)
It is not solution yet. It's not that easy. This will leave you at the integration below:
OK sorry... I'm easily satisfied. It looked so neat and tidy I thought it must be correct and finished.
10. dvs, I was stopped at H, I couldn't figure out how to solve H

Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 21, 2005
Today on TSR

TSR Pub Quiz 2018 - Anime

The first of our week-long series of entertainment quizzes

University open days

• University of Bradford
Wed, 21 Nov '18
• Buckinghamshire New University
Wed, 21 Nov '18
• Heriot-Watt University
Wed, 21 Nov '18
Poll
Useful resources

Make your revision easier

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

How to use LaTex

Writing equations the easy way

Study habits of A* students

Top tips from students who have already aced their exams

Create your own Study Planner

Never miss a deadline again

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE