Math Insight

Trigonometric substitution


This section continues development of relatively special tricks to do special kinds of integrals. Even though the application of such things is limited, it's nice to be aware of the possibilities, at least a little bit.

The key idea here is to use trig functions to be able to ‘take the square root’ in certain integrals. There are just three prototypes for the kind of thing we can deal with: $$\sqrt{1-x^2}\;\;\;\;\;\sqrt{1+x^2}\;\;\;\;\;\sqrt{x^2-1}$$ Examples will illustrate the point.

In rough terms, the idea is that in an integral where the ‘worst’ part is $\sqrt{1-x^2}$, replacing $x$ by $\sin u$ (and, correspondingly, $dx$ by $\cos u\;du$), we will be able to take the square root, and then obtain an integral in the variable $u$ which is one of the trigonometric integrals which in principle we now know how to do. The point is that then $$\sqrt{1-x^2}=\sqrt{1-\sin^2x}=\sqrt{\cos^2x}=\cos x$$ We have ‘taken the square root’.

For example, in $$\int \sqrt{1-x^2}\;dx$$ we replace $x$ by $\sin u$ and $dx$ by $\cos u\,du$ to obtain $$\int \sqrt{1-x^2}\;dx=\int \sqrt{1-\sin^2u}\;\cos u\;du= \int \sqrt{\cos^2u}\;\cos u\;du=$$ $$=\int\cos u \;\cos u\;du=\int\cos^2u\;du$$ Now we have an integral we know how to integrate: using the half-angle formula, this is $$\int\cos^2u\;du=\int{1+\cos 2u\over 2}\;du={u\over 2}+{\sin 2u\over 4}+C$$ And there still remains the issue of substituting back to obtain an expression in terms of $x$ rather than $u$. Since $x=\sin u$, it's just the definition of inverse function that $$u=\arcsin x$$ To express $\sin 2u$ in terms of $x$ is more aggravating. We use another half-angle formula $$\sin 2u=2\sin u\cos u$$ Then $${1\over 4}\sin 2u={1\over 4}\cdot 2\sin u\cos u={1\over 4}x\cdot\sqrt{1-x^2}$$ where ‘of course’ we used the Pythagorean identity to give us $$\cos u=\sqrt{1-\sin^2u}=\sqrt{1-x^2}$$ Whew.

The next type of integral we can ‘improve’ is one containing an expression $$\sqrt{1+x^2}$$ In this case, we use another Pythagorean identity $$1+\tan^2u=\sec^2u$$ (which we can get from the usual one $\cos^2u+\sin^2u=1$ by dividing by $\cos^2u$). So we'd let $$x=\tan u\;\;\;\;\;dx=\sec^2u\;du$$ (mustn't forget the $dx$ and $du$ business!).

For example, in $$\int { \sqrt{1+x^2} \over x }\;dx$$ we use $$x=\tan u\;\;\;\;\;dx=\sec^2u\;du$$ and turn the integral into \begin{align*}\int { \sqrt{1+x^2} \over x }\;dx&=\int { \sqrt{1+\tan^2 u} \over \tan u }\sec^2u\;du\\ &=\int { \sqrt{\sec^2u} \over \tan u }\sec^2u\;du\\ &= \int {\sec u\over \tan u}\sec^2u\;du=\int {1\over \sin u\cos^2u}\;du \end{align*} by rewriting everything in terms of $\cos u$ and $\sin u$.

For integrals containing $\sqrt{x^2-1}$, use $x=\sec u$ in order to invoke the Pythagorean identity $$\sec^2u-1=\tan^2u$$ so as to be able to ‘take the square root’. Let's not execute any examples of this, since nothing new really happens.

Rather,, let's examine some purely algebraic variants of these trigonometric substitutions, where we can get some mileage out of completing the square. For example, consider $$\int\sqrt{-2x-x^2}\;dx$$ The quadratic polynomial inside the square-root is not one of the three simple types we've looked at. But, by completing the square, we'll be able to rewrite it in essentially such forms: $$-2x-x^2=-(2x+x^2)=-(-1+1+2x+x^2)=-(-1+(1+x)^2)=1-(1+x)^2$$ Note that always when completing the square we ‘take out’ the coefficient in front of $x^2$ in order to see what's going on, and then put it back at the end.

So, in this case, we'd let $$\sin u=1+x\;\;\;\;\;\cos u\;du=dx$$

In another example, we might have $$\int\sqrt{8x-4x^2}\;dx$$ Completing the square again, we have $$8x-4x^2=-4(-2+x^2)=-4(-1+1-2x+x^2)=-4(-1+(x-1)^2)$$ Rather than put the whole ‘$-4$’ back, we only keep track of the $\pm$, and take a ‘$+4$’ outside the square root entirely: \begin{align*} \int\sqrt{8x-4x^2}\;dx&=\int\sqrt{-4(-1+(x-1)^2)}\;dx\\ &=2\int\sqrt{-(-1+(x-1)^2)}\;dx\\ &=2\int\sqrt{1-(x-1)^2)}\;dx \end{align*} Then we're back to a familiar situation.


  1. Tell what trig substitution to use for $\int x^8\sqrt{x^2-1}\,dx$
  2. Tell what trig substitution to use for $\int \sqrt{25+16x^2}\,dx$
  3. Tell what trig substitution to use for $\int \sqrt{1-x^2}\,dx$
  4. Tell what trig substitution to use for $\int \sqrt{9+4x^2}\,dx$
  5. Tell what trig substitution to use for $\int x^9\sqrt{x^2+1}\,dx$
  6. Tell what trig substitution to use for $\int x^8\sqrt{x^2-1}\,dx$