# 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(x)+sin^2(x)=1.

Multiplying both sides by a^2 gives a^2cos^2(x)+a^2sin^2(x)=a^2.

Dividing both sides of equation by cos^2(x) yields: a^2+a^2tan^2(x)=a^2sec^2(x).

From these equations we have that cos(x)=1/a sqrt(a^2-(asin(x))^2) and tan(x)=1/a sqrt((a sec(x))-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=asin(u).
2. If you have sqrt(x^2-a^2), make substitution x=a sec(u).
3. If you have sqrt(x^2+a^2), make substitution x=a tan(u).

However, carefully examine integral: maybe you don't need none of the above substitutions: for example, for integral int xsqrt(x^2+a^2)dx there is a simpler substitution u=x^2+a^2.

Example 1. Find int sqrt(4x^2-16)/x dx

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

Now, we are ready to make substitution x=2sec(u) , then dx=2sec(u)tan(u)du.

So, 2 int sqrt(x^2-4)/xdx=2 int sqrt(4 sec^2(u)-4)/(2sec(u)) 2sec(u)tan(u)du=

=4 int |tan(u)|tan(u)du

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

To return to old variable we will need to find tan(u) in terms of x. From our substitution we see that sec(u)=x/2.

Since tan(u)=sqrt(sec^2(u)-1) then tan(u)=sqrt((x^2)/4-1)=1/2 sqrt(x^2-4).

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

Finally, int (sqrt(4x^2-16))/xdx=4(1/2 sqrt(x^2-4)-arccos(2/x))+C=

=2(sqrt(x^2-4)-2arccos(2/x))+C.

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

Example 2. Calculate int_2^4 sqrt(4x^2-16)/xdx .

It is same integral as in example 1. First, simplify expression under square root: 2 int_2^4 sqrt(x^2-4)/x dx.

Let x=2sec(u) then dx=2sec(u)tan(u)du.

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

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

Now, integral becomes 4 int_0^(pi/3)tan^2(u)du=4 int_0^(pi/3)(sec^2(u)-1)du=4(tan(u)-u)|_0^(pi/2)=

=4((tan(0)-0)-(tan(pi/3)-(pi)/3))=-4(0-0-sqrt(3)+(pi)/3)=

=4sqrt(3)-(4pi)/3.

Example 3. Calculate int_(-4)^(-2) sqrt(4x^2-16)/xdx .

It is same integral as in example 1. First, simplify expression under square root: 2 int_(-4)^(-2) sqrt(x^2-4)/x dx .

Let x=2sec(u) then dx=2sec(u)tan(u)du.

x is changing from -4 to -2, so sec(u) is changing from (-4)/2=-2 to (-2)/2=-1, that's why cos(u) is changing from -1/2 to -1 and u is changing from (2pi)/3 to pi.

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

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

Now, integral becomes -4 int_((2pi)/3)^(pi)tan^2(u)du=-4 int_((2pi)/3)^(pi)(sec^2(u)-1)du=4(tan(u)-u)|_((2pi)/3)^(pi)=

=-4((tan(pi)-pi)-(tan((2pi)/3)-(2pi)/3))=-4(0-pi+sqrt(3)+(2pi)/3)=

=(4pi)/3-4sqrt(3).

Example 4. Calculate int 1/(x^2sqrt(4-x^2))dx .

We will need substitution 2 here: x=2sin(u) then dx=2cos(u)du and integral can be rewritten as:

int 1/(4sin^2(u)sqrt(4-4sin^2(u)))2cos(u)du=1/4 int 1/(sin^2(u)|cos(u)|) cos(u)du .

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

1/4 int 1/(sin^2(u))du=1/4 int csc^2(u)du=-1/4 cot(u)+C.

Since x=2sin(u) then sin(u)=x/2 and cos(u)=sqrt(1-(x^2)/4)=1/2 sqrt(4-x^2) , so cot(u)=cos(u)/sin(u)=(1/2 sqrt(4-x^2))/(1/2 x)=(sqrt(4-x^2))/x.

Finally, int 1/(x^2sqrt(4-x^2))dx=-(sqrt(4-x^2))/(4x)+C.

Example 5. Calculate int_0^2 (x^2)/(x^2+4)^(5/2)dx

Since (x^2+4)^(5/2)=(sqrt(x^2+4))^5 then make substitution number 3: x=2tan(u). In this case dx=2sec^2(u)du.

So, integral can be rewritten as int_0^2 (4tan^2(u))/(sqrt(4tan^2(u)+4))^5 2sec^2(u)du=int_0^2 (4tan^2(u))/(2|sec(u)|)^5 2sec^2(u)du.

Since x is changing from 0 to 2 then tan(u) is changing from 0/2=0 to 2/2=1. This means that u is changing from 0 to pi/4. This means that sec(u) is positive on this interval, so |sec(u)|=sec(u).

So, integral becomes 1/4 int_0^(pi/4)(tan^2(u))/(sec^3(u))du.

To calculate this integral it is better to convert it to sines and cosines: 1/4 int_0^(pi/4) (tan^2(u))/(sec^3(u))du=1/4 int_0^(pi/4) sin^2(u)cos(u)du .

Let t=sin(u) then dt=cos(u)du. Since u is changing from 0 to (pi)/4 then t is changing from sin(0)=0 to sin(pi/4)=1/sqrt(2).

And integral becomes 1/4int_0^(1/sqrt(2))t^2dt=1/12 t^3|_0^(1/sqrt(2))=1/12 (1/(sqrt(2))^3-0^3)=1/(24sqrt(2)).

Example 6. Evaluate int sqrt(x^2+4x+13)dx.

It seems that this integral can't be evaluated using above substitution. However, recall that any quadratic function ax^2+bx+c can be transformed into a(x+d)^2+e by completing a square.

So, x^2+4x+13=x^2+4x+4+9=(x+2)^2+9.

This is similar to sqrt(x^2+a^2) except x+2 term. But this doesn't matter. Let x+2=3tan(u). Then dx=3sec^2(u)du.

Integral, thus, becomes int sqrt(9tan^2(u)+9)*3sec^2(u)du=int 3|sec(u)|*3sec^2(u)du=9 int sec^3(u)du.

This integral was found in Trigonometric Integrals note: 9 int sec^3(u)du=9/2(sec(u)tan(u)+ln|sec(u)+tan(u)|)+C .

Since tan(u)=(x+2)/3 then sec(u)=sqrt(((x+2)^2)/9+1)=1/3sqrt(x^2+4x+13).

Finally, int sqrt(x^2+4x+13)dx=



=9/2(1/3sqrt(x^2+4x+13)*(x+2)/3+ln|(x+2)/3+sqrt(((x+2)/3)^2+1)|)+C=

=1/2(sqrt(x^2+4x+13)(x+2)+9text(arcsinh)((x+2)/3))+C.

Example 7. Evaluate int sqrt(e^x-1)dx .

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

Since e^(x/2)=sec(u) then 1/2sec(u)dx=sec(u)tan(u)du or dx=2tan(u)du.

Integral now becomes int sqrt(sec^2(u)-1)*2tan(u)du=2 int |tan(u)|tan(u)du.

Since integral is indefinite, we drop absolute value bars: 2 int tan^2(u)du=2 int (sec^2(u)-1)du=2(tan(u)-u)+C.

Since sec(u)=e^(x/2) then tan(u)=sqrt((e^(x/2))^2-1)=sqrt(e^x-1).

And u is either text(arcsec)(e^(x/2)) or arctan(sqrt(e^x-1)) (we choose second option).

So, int sqrt(e^x-1)dx=2(sqrt(e^x-1)-arctan(sqrt(e^x-1)))+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.