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Trigonometric Substitutions In Integrals

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Trigonometric Substitutions are especially useful when we want to get rid of $$$\sqrt{{{{x}}^{{2}}-{{a}}^{{2}}}}$$$, $$$\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}$$$ and $$$\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}$$$ under integral sign.

Recall that trignomeric identity states $$${{\cos}}^{{2}}{\left({x}\right)}+{{\sin}}^{{2}}{\left({x}\right)}={1}$$$.

Multiplying both sides by $$${{a}}^{{2}}$$$ gives $$${{a}}^{{2}}{{\cos}}^{{2}}{\left({x}\right)}+{{a}}^{{2}}{{\sin}}^{{2}}{\left({x}\right)}={{a}}^{{2}}$$$.

Dividing both sides of equation by $$${{\cos}}^{{2}}{\left({x}\right)}$$$ yields: $$${{a}}^{{2}}+{{a}}^{{2}}{{\tan}}^{{2}}{\left({x}\right)}={{a}}^{{2}}{{\sec}}^{{2}}{\left({x}\right)}$$$.

From these equations we have that $$${\cos{{\left({x}\right)}}}=\frac{{1}}{{a}}\sqrt{{{{a}}^{{2}}-{{\left({\operatorname{asin}{{\left({x}\right)}}}\right)}}^{{2}}}}$$$ and $$${\tan{{\left({x}\right)}}}=\frac{{1}}{{a}}\sqrt{{{\left({a}{\sec{{\left({x}\right)}}}\right)}-{{a}}^{{2}}}}$$$ (since integral is indefinite, we can drop absolute value bars, but this can't be done in definite integral). Notice similarity with above roots.

In general:

  1. If you have $$$\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}$$$, make substitution $$${x}={\operatorname{asin}{{\left({u}\right)}}}$$$.
  2. If you have $$$\sqrt{{{{x}}^{{2}}-{{a}}^{{2}}}}$$$, make substitution $$${x}={a}{\sec{{\left({u}\right)}}}$$$.
  3. If you have $$$\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}$$$, make substitution $$${x}={a}{\tan{{\left({u}\right)}}}$$$.

We already made first substitution when discussed Substitution Rule.

However, carefully examine integral: maybe you don't need none of the above substitutions: for example, for integral $$$\int{x}\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}{d}{x}$$$ there is a simpler substitution $$${u}={{x}}^{{2}}+{{a}}^{{2}}$$$.

Example 1. Find $$$\int\frac{\sqrt{{{4}{{x}}^{{2}}-{16}}}}{{x}}{d}{x}$$$

We see similar to above root in numerator, just need to rewrite it a bit: $$$\sqrt{{{4}{{x}}^{{2}}-{16}}}={2}\sqrt{{{{x}}^{{2}}-{4}}}$$$.

Now, we are ready to make substitution $$${x}={2}{\sec{{\left({u}\right)}}}$$$, then $$${d}{x}={2}{\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}{d}{u}$$$.

So, $$${2}\int\frac{\sqrt{{{{x}}^{{2}}-{4}}}}{{x}}{d}{x}={2}\int\frac{\sqrt{{{4}{{\sec}}^{{2}}{\left({u}\right)}-{4}}}}{{{2}{\sec{{\left({u}\right)}}}}}{2}{\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}{d}{u}=$$$

$$$={4}\int{\left|{\tan{{\left({u}\right)}}}\right|}{\tan{{\left({u}\right)}}}{d}{u}$$$

Since integral is indefinite we will drop absolute value bars: $$${4}\int{{\tan}}^{{2}}{\left({u}\right)}{d}{u}$$$. We know how to do such integrals (see Trigonometric Integrals note). So, $$${4}\int{{\tan}}^{{2}}{\left({u}\right)}{d}{u}={4}\int{\left({{\sec}}^{{2}}{\left({u}\right)}-{1}\right)}{d}{u}={4}{\left({\tan{{\left({u}\right)}}}-{u}\right)}+{C}$$$.

To return to old variable we will need to find $$${\tan{{\left({u}\right)}}}$$$ in terms of $$${x}$$$. From our substitution we see that $$${\sec{{\left({u}\right)}}}=\frac{{x}}{{2}}$$$.

Since $$${\tan{{\left({u}\right)}}}=\sqrt{{{{\sec}}^{{2}}{\left({u}\right)}-{1}}}$$$ then $$${\tan{{\left({u}\right)}}}=\sqrt{{\frac{{{{x}}^{{2}}}}{{4}}-{1}}}=\frac{{1}}{{2}}\sqrt{{{{x}}^{{2}}-{4}}}$$$.

Since $$${\sec{{\left({u}\right)}}}=\frac{{x}}{{2}}$$$ then $$${\cos{{\left({u}\right)}}}=\frac{{2}}{{x}}$$$ and $$${u}={\operatorname{arccos}{{\left(\frac{{2}}{{x}}\right)}}}$$$ (note that we could express $$${u}$$$ in terms of $$${x}$$$ through inverse secant: $$${u}=\text{arcsec}{\left(\frac{{x}}{{2}}\right)}$$$, people just more familiar with cosines therefore we use inverse cosine).

Finally, $$$\int\frac{{\sqrt{{{4}{{x}}^{{2}}-{16}}}}}{{x}}{d}{x}={4}{\left(\frac{{1}}{{2}}\sqrt{{{{x}}^{{2}}-{4}}}-{\operatorname{arccos}{{\left(\frac{{2}}{{x}}\right)}}}\right)}+{C}=$$$

$$$={2}{\left(\sqrt{{{{x}}^{{2}}-{4}}}-{2}{\operatorname{arccos}{{\left(\frac{{2}}{{x}}\right)}}}\right)}+{C}$$$.

Now, let's see how to handle absolute value bars in the definite integral.

Example 2. Calculate $$${\int_{{2}}^{{4}}}\frac{\sqrt{{{4}{{x}}^{{2}}-{16}}}}{{x}}{d}{x}$$$.

It is same integral as in example 1. First, simplify expression under square root: $$${2}{\int_{{2}}^{{4}}}\frac{\sqrt{{{{x}}^{{2}}-{4}}}}{{x}}{d}{x}$$$.

Let $$${x}={2}{\sec{{\left({u}\right)}}}$$$ then $$${d}{x}={2}{\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}{d}{u}$$$.

$$${x}$$$ is changing from 2 to 4, so $$${\sec{{\left({u}\right)}}}$$$ is changing from $$$\frac{{2}}{{2}}={1}$$$ to $$$\frac{{4}}{{2}}={2}$$$, that's why $$${\cos{{\left({u}\right)}}}$$$ is changing from $$$\frac{{1}}{{2}}$$$ to $$${1}$$$ and u is changing from 0 to $$$\frac{\pi}{{3}}$$$. Tangent on this interval is positive, so $$$\sqrt{{{{x}}^{{2}}-{4}}}=\sqrt{{{4}{{\sec}}^{{2}}{\left({u}\right)}-{4}}}={2}{\left|{\tan{{\left({u}\right)}}}\right|}={2}{\tan{{\left({u}\right)}}}$$$.

Note that in determining value of $$${u}$$$ we took the smallest positive value.

Now, integral becomes $$${4}{\int_{{0}}^{{\frac{\pi}{{3}}}}}{{\tan}}^{{2}}{\left({u}\right)}{d}{u}={4}{\int_{{0}}^{{\frac{\pi}{{3}}}}}{\left({{\sec}}^{{2}}{\left({u}\right)}-{1}\right)}{d}{u}={4}{\left({\tan{{\left({u}\right)}}}-{u}\right)}{{\mid}_{{0}}^{{\frac{\pi}{{2}}}}}=$$$

$$$={4}{\left({\left({\tan{{\left({0}\right)}}}-{0}\right)}-{\left({\tan{{\left(\frac{\pi}{{3}}\right)}}}-\frac{{\pi}}{{3}}\right)}\right)}=-{4}{\left({0}-{0}-\sqrt{{{3}}}+\frac{{\pi}}{{3}}\right)}=$$$

$$$={4}\sqrt{{{3}}}-\frac{{{4}\pi}}{{3}}$$$.

Example 3. Calculate $$${\int_{{-{4}}}^{{-{2}}}}\frac{\sqrt{{{4}{{x}}^{{2}}-{16}}}}{{x}}{d}{x}$$$.

It is same integral as in example 1. First, simplify expression under square root: $$${2}{\int_{{-{4}}}^{{-{2}}}}\frac{\sqrt{{{{x}}^{{2}}-{4}}}}{{x}}{d}{x}$$$.

Let $$${x}={2}{\sec{{\left({u}\right)}}}$$$ then $$${d}{x}={2}{\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}{d}{u}$$$.

$$${x}$$$ is changing from -4 to -2, so $$${\sec{{\left({u}\right)}}}$$$ is changing from $$$\frac{{-{4}}}{{2}}=-{2}$$$ to $$$\frac{{-{2}}}{{2}}=-{1}$$$, that's why $$${\cos{{\left({u}\right)}}}$$$ is changing from $$$-\frac{{1}}{{2}}$$$ to $$$-{1}$$$ and $$${u}$$$ is changing from $$$\frac{{{2}\pi}}{{3}}$$$ to $$$\pi$$$.

This means that tangent on this interval is negative, so $$$\sqrt{{{{x}}^{{2}}-{4}}}=\sqrt{{{4}{{\sec}}^{{2}}{\left({u}\right)}-{4}}}={2}{\left|{\tan{{\left({u}\right)}}}\right|}=-{2}{\tan{{\left({u}\right)}}}$$$.

Note that in determining value of $$${u}$$$ we took the smallest positive value.

Now, integral becomes $$$-{4}{\int_{{\frac{{{2}\pi}}{{3}}}}^{{\pi}}}{{\tan}}^{{2}}{\left({u}\right)}{d}{u}=-{4}{\int_{{\frac{{{2}\pi}}{{3}}}}^{{\pi}}}{\left({{\sec}}^{{2}}{\left({u}\right)}-{1}\right)}{d}{u}={4}{\left({\tan{{\left({u}\right)}}}-{u}\right)}{{\mid}_{{\frac{{{2}\pi}}{{3}}}}^{{\pi}}}=$$$

$$$=-{4}{\left({\left({\tan{{\left(\pi\right)}}}-\pi\right)}-{\left({\tan{{\left(\frac{{{2}\pi}}{{3}}\right)}}}-\frac{{{2}\pi}}{{3}}\right)}\right)}=-{4}{\left({0}-\pi+\sqrt{{{3}}}+\frac{{{2}\pi}}{{3}}\right)}=$$$

$$$=\frac{{{4}\pi}}{{3}}-{4}\sqrt{{{3}}}$$$.

Example 4. Calculate $$$\int\frac{{1}}{{{{x}}^{{2}}\sqrt{{{4}-{{x}}^{{2}}}}}}{d}{x}$$$.

We will need substitution 2 here: $$${x}={2}{\sin{{\left({u}\right)}}}$$$ then $$${d}{x}={2}{\cos{{\left({u}\right)}}}{d}{u}$$$ and integral can be rewritten as:

$$$\int\frac{{1}}{{{4}{{\sin}}^{{2}}{\left({u}\right)}\sqrt{{{4}-{4}{{\sin}}^{{2}}{\left({u}\right)}}}}}{2}{\cos{{\left({u}\right)}}}{d}{u}=\frac{{1}}{{4}}\int\frac{{1}}{{{{\sin}}^{{2}}{\left({u}\right)}{\left|{\cos{{\left({u}\right)}}}\right|}}}{\cos{{\left({u}\right)}}}{d}{u}$$$.

Since integral is indefinite, we can drop absolute value bars (we assume that cosine is positive):

$$$\frac{{1}}{{4}}\int\frac{{1}}{{{{\sin}}^{{2}}{\left({u}\right)}}}{d}{u}=\frac{{1}}{{4}}\int{{\csc}}^{{2}}{\left({u}\right)}{d}{u}=-\frac{{1}}{{4}}{\cot{{\left({u}\right)}}}+{C}$$$.

Now, we will need to return to old variables.

Since $$${x}={2}{\sin{{\left({u}\right)}}}$$$ then $$${\sin{{\left({u}\right)}}}=\frac{{x}}{{2}}$$$ and $$${\cos{{\left({u}\right)}}}=\sqrt{{{1}-\frac{{{{x}}^{{2}}}}{{4}}}}=\frac{{1}}{{2}}\sqrt{{{4}-{{x}}^{{2}}}}$$$, so $$${\cot{{\left({u}\right)}}}=\frac{{\cos{{\left({u}\right)}}}}{{\sin{{\left({u}\right)}}}}=\frac{{\frac{{1}}{{2}}\sqrt{{{4}-{{x}}^{{2}}}}}}{{\frac{{1}}{{2}}{x}}}=\frac{{\sqrt{{{4}-{{x}}^{{2}}}}}}{{x}}$$$.

Finally, $$$\int\frac{{1}}{{{{x}}^{{2}}\sqrt{{{4}-{{x}}^{{2}}}}}}{d}{x}=-\frac{{\sqrt{{{4}-{{x}}^{{2}}}}}}{{{4}{x}}}+{C}$$$.

Example 5. Calculate $$${\int_{{0}}^{{2}}}\frac{{{{x}}^{{2}}}}{{{\left({{x}}^{{2}}+{4}\right)}}^{{\frac{{5}}{{2}}}}}{d}{x}$$$

Since $$${{\left({{x}}^{{2}}+{4}\right)}}^{{\frac{{5}}{{2}}}}={{\left(\sqrt{{{{x}}^{{2}}+{4}}}\right)}}^{{5}}$$$ then make substitution number 3: $$${x}={2}{\tan{{\left({u}\right)}}}$$$. In this case $$${d}{x}={2}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}$$$.

So, integral can be rewritten as $$${\int_{{0}}^{{2}}}\frac{{{4}{{\tan}}^{{2}}{\left({u}\right)}}}{{{\left(\sqrt{{{4}{{\tan}}^{{2}}{\left({u}\right)}+{4}}}\right)}}^{{5}}}{2}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}={\int_{{0}}^{{2}}}\frac{{{4}{{\tan}}^{{2}}{\left({u}\right)}}}{{{\left({2}{\left|{\sec{{\left({u}\right)}}}\right|}\right)}}^{{5}}}{2}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}$$$.

Since $$${x}$$$ is changing from 0 to 2 then $$${\tan{{\left({u}\right)}}}$$$ is changing from $$$\frac{{0}}{{2}}={0}$$$ to $$$\frac{{2}}{{2}}={1}$$$. This means that $$${u}$$$ is changing from $$${0}$$$ to $$$\frac{\pi}{{4}}$$$. This means that $$${\sec{{\left({u}\right)}}}$$$ is positive on this interval, so $$${\left|{\sec{{\left({u}\right)}}}\right|}={\sec{{\left({u}\right)}}}$$$.

So, integral becomes $$$\frac{{1}}{{4}}{\int_{{0}}^{{\frac{\pi}{{4}}}}}\frac{{{{\tan}}^{{2}}{\left({u}\right)}}}{{{{\sec}}^{{3}}{\left({u}\right)}}}{d}{u}$$$.

To calculate this integral it is better to convert it to sines and cosines: $$$\frac{{1}}{{4}}{\int_{{0}}^{{\frac{\pi}{{4}}}}}\frac{{{{\tan}}^{{2}}{\left({u}\right)}}}{{{{\sec}}^{{3}}{\left({u}\right)}}}{d}{u}=\frac{{1}}{{4}}{\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\sin}}^{{2}}{\left({u}\right)}{\cos{{\left({u}\right)}}}{d}{u}$$$.

Let $$${t}={\sin{{\left({u}\right)}}}$$$ then $$${d}{t}={\cos{{\left({u}\right)}}}{d}{u}$$$. Since u is changing from 0 to $$$\frac{{\pi}}{{4}}$$$ then $$${t}$$$ is changing from sin(0)=0 to $$${\sin{{\left(\frac{\pi}{{4}}\right)}}}=\frac{{1}}{\sqrt{{{2}}}}$$$.

And integral becomes $$$\frac{{1}}{{4}}{\int_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}{{t}}^{{2}}{d}{t}=\frac{{1}}{{12}}{{t}}^{{3}}{{\mid}_{{0}}^{{\frac{{1}}{\sqrt{{{2}}}}}}}=\frac{{1}}{{12}}{\left(\frac{{1}}{{{\left(\sqrt{{{2}}}\right)}}^{{3}}}-{{0}}^{{3}}\right)}=\frac{{1}}{{{24}\sqrt{{{2}}}}}$$$.

Example 6. Evaluate $$$\int\sqrt{{{{x}}^{{2}}+{4}{x}+{13}}}{d}{x}$$$.

It seems that this integral can't be evaluated using above substitution. However, recall that any quadratic function $$${a}{{x}}^{{2}}+{b}{x}+{c}$$$ can be transformed into $$${a}{{\left({x}+{d}\right)}}^{{2}}+{e}$$$ by completing a square.

So, $$${{x}}^{{2}}+{4}{x}+{13}={{x}}^{{2}}+{4}{x}+{4}+{9}={{\left({x}+{2}\right)}}^{{2}}+{9}$$$.

This is similar to $$$\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}$$$ except $$${x}+{2}$$$ term. But this doesn't matter. Let $$${x}+{2}={3}{\tan{{\left({u}\right)}}}$$$. Then $$${d}{x}={3}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}$$$.

Integral, thus, becomes $$$\int\sqrt{{{9}{{\tan}}^{{2}}{\left({u}\right)}+{9}}}\cdot{3}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}=\int{3}{\left|{\sec{{\left({u}\right)}}}\right|}\cdot{3}{{\sec}}^{{2}}{\left({u}\right)}{d}{u}={9}\int{{\sec}}^{{3}}{\left({u}\right)}{d}{u}$$$.

This integral was found in Trigonometric Integrals note: $$${9}\int{{\sec}}^{{3}}{\left({u}\right)}{d}{u}=\frac{{9}}{{2}}{\left({\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}+{\ln}{\left|{\sec{{\left({u}\right)}}}+{\tan{{\left({u}\right)}}}\right|}\right)}+{C}$$$.

Since $$${\tan{{\left({u}\right)}}}=\frac{{{x}+{2}}}{{3}}$$$ then $$${\sec{{\left({u}\right)}}}=\sqrt{{\frac{{{{\left({x}+{2}\right)}}^{{2}}}}{{9}}+{1}}}=\frac{{1}}{{3}}\sqrt{{{{x}}^{{2}}+{4}{x}+{13}}}$$$.

Finally, $$$\int\sqrt{{{{x}}^{{2}}+{4}{x}+{13}}}{d}{x}=$$$

$$$=\frac{{9}}{{2}}{\left(\frac{{1}}{{3}}\sqrt{{{{x}}^{{2}}+{4}{x}+{13}}}\cdot\frac{{{x}+{2}}}{{3}}+{\ln}{\left|\frac{{{x}+{2}}}{{3}}+\sqrt{{{{\left(\frac{{{x}+{2}}}{{3}}\right)}}^{{2}}+{1}}}\right|}\right)}+{C}=$$$

$$$=\frac{{1}}{{2}}{\left(\sqrt{{{{x}}^{{2}}+{4}{x}+{13}}}{\left({x}+{2}\right)}+{9}\text{arcsinh}{\left(\frac{{{x}+{2}}}{{3}}\right)}\right)}+{C}$$$.

Example 7. Evaluate $$$\int\sqrt{{{{e}}^{{x}}-{1}}}{d}{x}$$$.

Again expression under the square root is not like the above 3 expressions. However, note that $$$\sqrt{{{{e}}^{{x}}-{1}}}=\sqrt{{{{\left({{e}}^{{\frac{{x}}{{2}}}}\right)}}^{{2}}-{1}}}$$$. And it is similar to substitution 2. So, let $$${{e}}^{{\frac{{x}}{{2}}}}={\sec{{\left({u}\right)}}}$$$ then $$$\frac{{1}}{{2}}{{e}}^{{\frac{{x}}{{2}}}}{d}{x}={\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}{d}{u}$$$. We need to express $$${d}{x}$$$ in terms of $$${u}$$$ only.

Since $$${{e}}^{{\frac{{x}}{{2}}}}={\sec{{\left({u}\right)}}}$$$ then $$$\frac{{1}}{{2}}{\sec{{\left({u}\right)}}}{d}{x}={\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}{d}{u}$$$ or $$${d}{x}={2}{\tan{{\left({u}\right)}}}{d}{u}$$$.

Integral now becomes $$$\int\sqrt{{{{\sec}}^{{2}}{\left({u}\right)}-{1}}}\cdot{2}{\tan{{\left({u}\right)}}}{d}{u}={2}\int{\left|{\tan{{\left({u}\right)}}}\right|}{\tan{{\left({u}\right)}}}{d}{u}$$$.

Since integral is indefinite, we drop absolute value bars: $$${2}\int{{\tan}}^{{2}}{\left({u}\right)}{d}{u}={2}\int{\left({{\sec}}^{{2}}{\left({u}\right)}-{1}\right)}{d}{u}={2}{\left({\tan{{\left({u}\right)}}}-{u}\right)}+{C}$$$.

Since $$${\sec{{\left({u}\right)}}}={{e}}^{{\frac{{x}}{{2}}}}$$$ then $$${\tan{{\left({u}\right)}}}=\sqrt{{{{\left({{e}}^{{\frac{{x}}{{2}}}}\right)}}^{{2}}-{1}}}=\sqrt{{{{e}}^{{x}}-{1}}}$$$.

And $$${u}$$$ is either $$$\text{arcsec}{\left({{e}}^{{\frac{{x}}{{2}}}}\right)}$$$ or $$${\operatorname{arctan}{{\left(\sqrt{{{{e}}^{{x}}-{1}}}\right)}}}$$$ (we choose second option).

So, $$$\int\sqrt{{{{e}}^{{x}}-{1}}}{d}{x}={2}{\left(\sqrt{{{{e}}^{{x}}-{1}}}-{\operatorname{arctan}{{\left(\sqrt{{{{e}}^{{x}}-{1}}}\right)}}}\right)}+{C}$$$.

Note, that this integral can be solved another way: with double substitution; first substitution is $$${u}={{e}}^{{x}}$$$ and second is $$${t}=\sqrt{{{u}-{1}}}$$$.

We have seen (last two examples) that some integrals can be converted into integrals that can be solved using trigonometric substitution described above.