Tangent Substitution
This is a great technique that introduces tangent into an integral which originally doesn't have any trig. This makes integration easier and sometimes possible where it wasn't before.
Make sure you have throughly gone through the trig substitution page since this page builds on that discussion.
Recommended Books on Amazon (affiliate links)  

Once you have determined that you have an integrand containing an \(x^2 + a^2\) term, you use these equations for the substitution.
Tangent Substitution  

\(a\) is a constant; \(x\) and \(\theta\) are variables  
integrand  \(x^2 + a^2\) 
substitution  \(x = a\tan(\theta)\) 
identity  \( 1+\tan^2(\theta) = \sec^2(\theta)\) 
Notice that when we perform the substitution, we have
\(\begin{array}{rcl}
a^2+x^2 & = & a^2 + (a\tan(\theta))^2 \\
& = & a^2 + a^2 \tan^2(\theta) \\
& = & a^2 (1 + \tan^2(\theta)) \\
& = & a^2 \sec^2(\theta) \\
\end{array}\)
In the last equation we end up with two squared terms, which we can easily take the square root of, for example, to simplify.
As mentioned in step 2 above, we need to build a triangle based on our choice of substitution, in this case
\(x = a\tan(\theta) \to \tan(\theta) = x/a\). So, we draw a right triangle, select one of the other two angles to be \(\theta\) and label \(a\) and \(x\) accordingingly. Then we use the Pythagorean Theorem to get the expression for the other side. This gives us the triangle on the right.
Next, we substitute in the integral for \(x = a\tan(\theta))\) and \(dx = a\sec^2(\theta)~d\theta \). This last substitution is important to remember since we can't just replace \(dx\) with \(d\theta\) and it is a common source of error for many students.
Finally, once we are done integrating, we use the same triangle above to convert all the terms back into \(x\). It is not correct to leave \(\theta\)'s in the final answer.
Here is a great video with all this laid out in one problem.
Comment On Notation  Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.
video by Integrals ForYou 

Okay, time for some practice problems.
Practice
Evaluate these integrals using tangent substitution. Give your answers in exact, simplified and factored form.
Basic
\(\displaystyle{ \int{\frac{dx}{(x^2+1)^2} } }\)
Problem Statement 

Evaluate \(\displaystyle{ \int{\frac{dx}{(x^2+1)^2} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Final Answer 

\(\displaystyle{ \int{\frac{dx}{(x^2+1)^2} } = \frac{1}{2} \left( \arctan(x) + \frac{x}{x^2+1} \right) + C }\)
Problem Statement
Evaluate \(\displaystyle{ \int{\frac{dx}{(x^2+1)^2} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Solution
video by Krista King Math 

Final Answer
\(\displaystyle{ \int{\frac{dx}{(x^2+1)^2} } = \frac{1}{2} \left( \arctan(x) + \frac{x}{x^2+1} \right) + C }\)
Log in to rate this practice problem and to see it's current rating. 

\(\displaystyle{ \int_{0}^{10}{ \frac{dx}{x^2+25} } }\)
Problem Statement 

Evaluate \(\displaystyle{ \int_{0}^{10}{ \frac{dx}{x^2+25} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Final Answer 

\(\displaystyle{ \int_{0}^{10}{\frac{dx}{x^2+25} } = \frac{1}{5}\arctan(2) }\)
Problem Statement
Evaluate \(\displaystyle{ \int_{0}^{10}{ \frac{dx}{x^2+25} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Solution
In this video, she makes a major mistake in notation when working definite integrals using substitution. She carries along the limits of integration for the x variable when she has an integral using θ. This is incorrect and could lead to major errors and loss of points on homework and exams. She either should drop the limits of integration while she is using θ or she should convert the limits in terms of θ. The second way is recommended since she would not have to convert the result of the integral back to x before substituting.
For converting, we use the equation \(x=5\tan(\theta)\) and convert \(x=0 \to \theta = 0\) and \(x=10 \to \theta = \arctan(2) \)
\(\displaystyle{\int_{0}^{10}{ \frac{5\sec^2 \theta}{(5\tan\theta)^2 + 25} d\theta }}\) should be \(\displaystyle{ \int_{0}^{\arctan(2)}{ \frac{5\sec^2 \theta}{(5\tan\theta)^2 + 25} d\theta } }\)
video by Krista King Math 

Final Answer
\(\displaystyle{ \int_{0}^{10}{\frac{dx}{x^2+25} } = \frac{1}{5}\arctan(2) }\)
Log in to rate this practice problem and to see it's current rating. 

\(\displaystyle{ \int_{0}^{2}{ \frac{dx}{\sqrt{x^2+4}} } }\)
Problem Statement 

Evaluate \(\displaystyle{ \int_{0}^{2}{ \frac{dx}{\sqrt{x^2+4}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Final Answer 

\(\displaystyle{ \int_{0}^{2}{ \frac{dx}{\sqrt{x^2+4}} } = \ln \left\sqrt{2}+1\right }\)
Problem Statement
Evaluate \(\displaystyle{ \int_{0}^{2}{ \frac{dx}{\sqrt{x^2+4}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Solution
video by PatrickJMT 

Final Answer
\(\displaystyle{ \int_{0}^{2}{ \frac{dx}{\sqrt{x^2+4}} } = \ln \left\sqrt{2}+1\right }\)
Log in to rate this practice problem and to see it's current rating. 

\(\displaystyle{ \int{ \frac{dx}{(a^2+x^2)^{3/2}} } }\)
Problem Statement
Evaluate \(\displaystyle{ \int{ \frac{dx}{(a^2+x^2)^{3/2}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Solution
video by Dr Chris Tisdell 

Log in to rate this practice problem and to see it's current rating. 

Intermediate
\(\displaystyle{ \int{ \sqrt{x^2+4} ~dx } }\)
Problem Statement 

Evaluate \(\displaystyle{ \int{ \sqrt{x^2+4} ~dx } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Final Answer 

\(\displaystyle{ \int{ \sqrt{x^2+4} ~dx } = \frac{x\sqrt{x^2+4}}{2} + }\) \(\displaystyle{ 2\ln\left \frac{x+\sqrt{x^2+4}}{2} \right + C }\)
Problem Statement
Evaluate \(\displaystyle{ \int{ \sqrt{x^2+4} ~dx } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Solution
video by Krista King Math 

Final Answer
\(\displaystyle{ \int{ \sqrt{x^2+4} ~dx } = \frac{x\sqrt{x^2+4}}{2} + }\) \(\displaystyle{ 2\ln\left \frac{x+\sqrt{x^2+4}}{2} \right + C }\)
Log in to rate this practice problem and to see it's current rating. 

\(\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+1}} ~ dx } }\)
Problem Statement
\(\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+1}} ~ dx } }\)
Solution
video by Michael Penn 

Log in to rate this practice problem and to see it's current rating. 

\(\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+9}} ~dx } }\)
Problem Statement 

Evaluate \(\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+9}} ~dx } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Final Answer 

\(\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+9}} ~dx } = \frac{1}{3}(x^218)\sqrt{x^2+9} + C }\)
Problem Statement
Evaluate \(\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+9}} ~dx } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Solution
This problem is solved in two consecutive videos.
video by PatrickJMT 

video by PatrickJMT 

Final Answer
\(\displaystyle{ \int{ \frac{x^3}{\sqrt{x^2+9}} ~dx } = \frac{1}{3}(x^218)\sqrt{x^2+9} + C }\)
Log in to rate this practice problem and to see it's current rating. 

\(\displaystyle{ \int{ \frac{dx}{a+bx^2} } }\)
Problem Statement
Evaluate \(\displaystyle{ \int{ \frac{dx}{a+bx^2} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Solution
video by Michel vanBiezen 

Log in to rate this practice problem and to see it's current rating. 

Advanced
\(\displaystyle{ \int{ \frac{x}{\sqrt{x^2+2x+2}} ~ dx } }\)
Problem Statement
\(\displaystyle{ \int{ \frac{x}{\sqrt{x^2+2x+2}} ~ dx } }\)
Solution
video by Michael Penn 

Log in to rate this practice problem and to see it's current rating. 

Expanding The Use of The Tangent Substitution
In the discussion so far on this page, we have assumed that the \(x\) variable term is always exactly \(x\). There is a generalization of this where you might have \([f(x)]^2 + a^2\) and \(f(x) \neq x\). This technique still works and you just change the formula as follows.
Expanded Tangent Substitution  

integrand 
substitution  
\([f(x)]^2+a^2\) 
\(f(x) = a\tan(\theta)\)  
Example  
\(e^{2x}+9\) 
\(e^x = 3\tan(\theta)\) 
\(f(x)\) can be pretty much anything except for a constant. So check that it contains at least one variable. In the example above, \(f(x)=e^x\).
Note: The integrand may not be given in the exact form \( [f(x)]^2 + a^2 \). You may need to complete the square in order to get this form.
Okay, let's work some practice problems using these ideas. You don't want to work these problems until you have mastered the basic technique in the previous section on this page.
Practice
Evaluate these integrals using tangent substitution. Give your answers in exact, simplified and factored form.
\(\displaystyle{ \int{ \sqrt{9+16x^2} ~dx } }\)
Problem Statement 

Evaluate \(\displaystyle{ \int{ \sqrt{9+16x^2} ~dx } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Final Answer 

\(\displaystyle{ \frac{x\sqrt{9+16x^2}}{2} + \frac{9}{8}\ln\left 4x + \sqrt{9+16x^2} \right + C}\)
Problem Statement
Evaluate \(\displaystyle{ \int{ \sqrt{9+16x^2} ~dx } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Solution
During the course of this problem, she drops a \(9/4\) term from her work. So the answer at the end of the video should be multiplied by \(9/4\) to get the correct answer. [She puts a note at about the 14:22 mark that points this out.] We give the correct answer.
video by Krista King Math 

Final Answer
\(\displaystyle{ \frac{x\sqrt{9+16x^2}}{2} + \frac{9}{8}\ln\left 4x + \sqrt{9+16x^2} \right + C}\)
Log in to rate this practice problem and to see it's current rating. 

\(\displaystyle{ \int{ \frac{dx}{\sqrt{9x^2+4}} } }\)
Problem Statement 

Evaluate \(\displaystyle{ \int{ \frac{dx}{\sqrt{9x^2+4}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Final Answer 

\(\displaystyle{ \int{ \frac{dx}{\sqrt{9x^2+4}} } = \frac{1}{3} \ln\left \sqrt{9x^2+4}+3x \right + C }\)
Problem Statement
Evaluate \(\displaystyle{ \int{ \frac{dx}{\sqrt{9x^2+4}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Solution
Note: The answer we show here is not the same as the answer that he shows in his video. However, they are both correct. Here is why.
His answer is \(\displaystyle{ \frac{1}{3}\ln\left \frac{\sqrt{9x^2+4}}{2} + \frac{3x}{2}\right + C }\).
We simplified this as follows.
Looking at the right side of the equation, we have a common denominator. 
\(\displaystyle{ \frac{1}{3}\ln\left \frac{\sqrt{9x^2+4}}{2} + \frac{3x}{2}\right+C}\) 
\(\displaystyle{ \frac{1}{3}\ln\left \frac{\sqrt{9x^2+4} + 3x}{2}\right+C}\) 
Next, we use the logarithm rule \( \ln(a/b) = \ln a  \ln b\). 
\(\displaystyle{ \frac{1}{3} \left[ \ln\left \sqrt{9x^2+4} + 3x \right  \ln2 \right] +C}\) 
Now we have a constant \(\ln(2)/3\) that we absorb into the constant C to get our final answer. 
In these steps, we do what many teachers do and just absorb constants into the unknown constant \(C\). We admit that this is not very good notation but most teachers do it. So you need to get used to it. Technically, we should say \(C_1 = C\ln(2)/3\) and write our answer as \(\displaystyle{\int{\frac{dx}{\sqrt{9x^2+4}}}= \frac{1}{3}\ln\left \sqrt{9x^2+4} + 3x\right+C_1}\). This would make it clear that our constant \(C_1\) is different than his constant \(C\) in the video. 
video by PatrickJMT 

Final Answer
\(\displaystyle{ \int{ \frac{dx}{\sqrt{9x^2+4}} } = \frac{1}{3} \ln\left \sqrt{9x^2+4}+3x \right + C }\)
Log in to rate this practice problem and to see it's current rating. 

\(\displaystyle{ \int{ \frac{dx}{(4x^2+9)^2} } }\)
Problem Statement 

Evaluate \(\displaystyle{ \int{ \frac{dx}{(4x^2+9)^2} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Final Answer 

\(\displaystyle{ \frac{1}{108} \left[ \arctan(2x/3) + \frac{6x}{4x^2+9} \right] + C }\)
Problem Statement
Evaluate \(\displaystyle{ \int{ \frac{dx}{(4x^2+9)^2} } }\) using trig substitution. Give your answer in exact, simplified, factored form.
Solution
substitution: \(2x = 3\tan \theta \)
related equations: \( 2dx = 3\sec^2 \theta ~d\theta \) and \( 4x^2 = 9\tan^2 \theta \)
\(\displaystyle{ \int{ \frac{dx}{(4x^2+9)^2} } }\) 
\(\displaystyle{ \int{ \frac{(3/2)\sec^2 \theta ~d\theta}{(9\tan^2 \theta + 9)^2} } }\) 
\(\displaystyle{ \frac{3}{2}\int{ \frac{\sec^2 \theta ~d\theta}{81(\tan^2 \theta + 1)^2} } }\) 
\(\displaystyle{\frac{1}{54} \int{ \frac{\sec^2 \theta ~d\theta}{\sec^4 \theta} } }\) 
\(\displaystyle{ \frac{1}{54} \int{ \cos^2 \theta ~d\theta } }\) 
\(\displaystyle{ \frac{1}{108} \int{ 1 + \cos(2\theta) ~d\theta } }\) 
\(\displaystyle{ \frac{1}{108} \left[ \theta + \frac{\sin(2\theta)}{2} \right] + C }\) 
\(\displaystyle{ \frac{1}{108} \left[ \theta + \sin \theta \cos \theta \right] + C }\) 
\(\displaystyle{ \frac{1}{108} \left[ \arctan(2x/3) + \frac{6x}{4x^2+9} \right] + C }\) 
Final Answer
\(\displaystyle{ \frac{1}{108} \left[ \arctan(2x/3) + \frac{6x}{4x^2+9} \right] + C }\)
Log in to rate this practice problem and to see it's current rating. 

Solve the initial value problem \(\displaystyle{ \frac{dy}{dt} = \frac{1}{25+9t^2}, y(5/3) = \frac{\pi}{30} }\)
Problem Statement 

Solve the initial value problem \(\displaystyle{ \frac{dy}{dt} = \frac{1}{25+9t^2}, y(5/3) = \frac{\pi}{30} }\)
Hint 

If you don't know how to work initial value problems, you can still follow the trig substitution part of this video.
Problem Statement 

Solve the initial value problem \(\displaystyle{ \frac{dy}{dt} = \frac{1}{25+9t^2}, y(5/3) = \frac{\pi}{30} }\)
Final Answer 

\(\displaystyle{ y(t) = \frac{1}{15} \arctan\left(\frac{3t}{5}\right) + \frac{\pi}{60} }\)
Problem Statement
Solve the initial value problem \(\displaystyle{ \frac{dy}{dt} = \frac{1}{25+9t^2}, y(5/3) = \frac{\pi}{30} }\)
Hint
If you don't know how to work initial value problems, you can still follow the trig substitution part of this video.
Solution
video by Krista King Math 

Final Answer
\(\displaystyle{ y(t) = \frac{1}{15} \arctan\left(\frac{3t}{5}\right) + \frac{\pi}{60} }\)
Log in to rate this practice problem and to see it's current rating. 

Now, let's look at the next substitution technique, sine substitution.
Really UNDERSTAND Calculus
Log in to rate this page and to see it's current rating.
Topics You Need To Understand For This Page 

To bookmark this page and practice problems, log in to your account or set up a free account.
Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
 
free ideas to save on books 

I recently started a Patreon account to help defray the expenses associated with this site. To keep this site free, please consider supporting me. 

Support 17Calculus on Patreon 
next: sine substitution → 



Practice Instructions
Evaluate these integrals using tangent substitution. Give your answers in exact, simplified and factored form.