##### Integration of a inverse Trigtonometric Forms

We know the derivatives of the inverse trigonometric functions

$\displaystyle (\text{arcsinx}{)}’=\frac{1}{{\sqrt{{1-{{x}^{2}}}}}}$

$\displaystyle (\text{arccosx}{)}’=\frac{{-1}}{{\sqrt{{1-{{x}^{2}}}}}}$

$\displaystyle (arctgx{)}’=\frac{1}{{1+{{x}^{2}}}}$

$\displaystyle (\text{arccotgx}{)}’=\frac{{-1}}{{1+{{x}^{2}}}}$

$\displaystyle (\text{arcsecx}{)}’=\frac{1}{{\left| x \right|\sqrt{{{{x}^{2}}-1}}}}$

$\displaystyle (\text{arccscx}{)}’=\frac{{-1}}{{\left| x \right|\sqrt{{{{x}^{2}}-1}}}}$

Using those derivatives above, we can obtain the integrals as below, where is a function of that u=f(x).

$\displaystyle \int{{\frac{{du}}{{\sqrt{{{{a}^{2}}-{{u}^{2}}}}}}}}=\arcsin \frac{u}{a}+C$

$\displaystyle \int{{\frac{{du}}{{{{a}^{2}}+{{u}^{2}}}}}}=\frac{1}{a}\arctan \frac{u}{a}+C$

$\displaystyle \int{{\frac{{du}}{{u\sqrt{{{{u}^{2}}-{{a}^{2}}}}}}}}$$\displaystyle =\frac{1}{a}\operatorname{arcsec}\frac{{\left| u \right|}}{a}+C Example 1: Find the integral \displaystyle \int{{\frac{{dx}}{{\sqrt{{36-{{x}^{2}}}}}}}}? Applying the first formula above \displaystyle \int{{\frac{{du}}{{\sqrt{{{{a}^{2}}-{{u}^{2}}}}}}}}=\arcsin \frac{u}{a}+C we get: \displaystyle \int{{\frac{{dx}}{{\sqrt{{36-{{x}^{2}}}}}}}}=\int{{\frac{{dx}}{{\sqrt{{{{{(6)}}^{2}}-{{x}^{2}}}}}}}}=\arcsin \frac{x}{6}+C You can also find the answer as: \displaystyle si{{n}^{{-1}}}\left( {\frac{x}{6}} \right)+C so don’t get confused it’s the same thing. Example 2: Find the integral \displaystyle \int{{\frac{{3dx}}{{\sqrt{{9-9{{x}^{2}}}}}}}} On a first look this integral doesn’t resemble with any of the two forms above. But using u-substitution we can transform it into one of the forms. Let \displaystyle u=3x then \displaystyle du=3dx Our integral becomes, \displaystyle \int{{\frac{{3dx}}{{\sqrt{{9-9{{x}^{2}}}}}}}}=\int{{\frac{{du}}{{\sqrt{{9-{{u}^{2}}}}}}}} Now we can write it as the second form above and find the integral, \displaystyle \int{{\frac{{du}}{{\sqrt{{{{{(3)}}^{2}}-{{u}^{2}}}}}}}}=\arcsin \frac{u}{3}+C Then substitute back the u \displaystyle \int{{\frac{{3dx}}{{\sqrt{{9-9{{x}^{2}}}}}}}}$$\displaystyle =\arcsin \frac{{3x}}{3}+C$$\displaystyle =\arcsin x+C Another simple method in this case is: You can factorize 9 inside the root and then simplify it with three after finding the root \displaystyle \int{{\frac{{3dx}}{{\sqrt{{9-9{{x}^{2}}}}}}}}=\int{{\frac{{3dx}}{{\sqrt{{9(1-{{x}^{2}})}}}}=}} \displaystyle \int{{\frac{{3dx}}{{3\sqrt{{(1-{{x}^{2}})}}}}}}=\int{{\frac{{dx}}{{\sqrt{{(1-{{x}^{2}})}}}}}} Now we see that we have obtained directly the second form above and we find the integral \displaystyle \int{{\frac{{dx}}{{\sqrt{{(1-{{x}^{2}})}}}}}}=\text{arcsinx}+C But this doesn’t work every time so the more general method is by using the u-substitution method. Let’s see another example when we can’t factorize. Example 3: Find the integral \displaystyle \int{{\frac{{4dx}}{{\sqrt{{25-16{{x}^{2}}}}}}}} On a first look this integral doesn’t resemble with any of the two forms above. But using u-substitution we can transform it into one of the forms. Let \displaystyle u=4x then \displaystyle du=4dx Our integral becomes, \displaystyle \int{{\frac{{4dx}}{{\sqrt{{25-16{{x}^{2}}}}}}}}=\int{{\frac{{du}}{{\sqrt{{25-{{{(u)}}^{2}}}}}}}} Now we can write it as the second form above and find the integral, \displaystyle \int{{\frac{{du}}{{\sqrt{{{{{(5)}}^{2}}-{{{(u)}}^{2}}}}}}}}=\arcsin \frac{u}{5}+C Then substitute back the u \displaystyle \int{{\frac{{4dx}}{{\sqrt{{25-16{{x}^{2}}}}}}}}=\arcsin \frac{{4x}}{5}+C Example 4: Find the integral \displaystyle \int{{\frac{{dx}}{{9+4{{x}^{2}}}}}}? On a first look this integral doesn’t resemble with any of the two forms above. After a transformation we see that we have the first form above \displaystyle \int{{\frac{{dx}}{{9+4{{x}^{2}}}}}}=\int{{\frac{{dx}}{{{{{(3)}}^{2}}+{{{(2x)}}^{2}}}}}}= Now finding the integral \displaystyle \int{{\frac{{dx}}{{{{{(3)}}^{2}}+{{{(2x)}}^{2}}}}}}=\frac{1}{3}\arctan \frac{{2x}}{3}+C Example 5: Find the integral \displaystyle \int{{\frac{{3dx}}{{{{x}^{2}}+4x+5}}}}? On a first look this integral doesn’t resemble with any of the two forms above. But let’s transform the denominator \displaystyle {{x}^{2}}+4x+5=({{x}^{2}}+4x+4)+1={{(x+2)}^{2}}+1 After rewriting the integral we see that we have the first form above \displaystyle \int{{\frac{{3dx}}{{{{x}^{2}}+4x+5}}}}=\int{{\frac{{3dx}}{{{{{(x+2)}}^{2}}+1}}}} Now to find the integral we use the u-substitution method Let \displaystyle u=x+2 then \displaystyle du=dx Substituting and finding the integral \displaystyle \int{{\frac{{3dx}}{{{{{(x+2)}}^{2}}+1}}}}=3\int{{\frac{{du}}{{{{u}^{2}}+1}}}}=3\arctan u+C Then substitute back the u \displaystyle \int{{\frac{{3dx}}{{{{{(x+2)}}^{2}}+1}}}}=3\arctan (x+2)+C Example 6: Find the integral \displaystyle \int{{\frac{{dx}}{{x\sqrt{{16{{x}^{2}}-9}}}}}}. On a first look this integral doesn’t resemble with any of the two forms above. But let’s transform it into the third form \displaystyle \int{{\frac{{dx}}{{x\sqrt{{16{{x}^{2}}-9}}}}}}=\int{{\frac{{4dx}}{{4x\sqrt{{{{{(4x)}}^{2}}-{{{(3)}}^{2}}}}}}}} Now we use the u-substitution method Let \displaystyle u=4x then \displaystyle du=4dx Substituting and finding the integral \displaystyle \int{{\frac{{4dx}}{{4x\sqrt{{{{{(4x)}}^{2}}-{{{(3)}}^{2}}}}}}}}$$\displaystyle =\int{{\frac{{du}}{{u\sqrt{{{{u}^{2}}-{{3}^{2}}}}}}}}$$\displaystyle =\frac{1}{3}\operatorname{arcsec}\frac{{\left| u \right|}}{3}+C Then substitute back the u \displaystyle \int{{\frac{{4dx}}{{4x\sqrt{{{{{(4x)}}^{2}}-{{{(3)}}^{2}}}}}}}}$$\displaystyle =\frac{1}{3}\operatorname{arcsec}\frac{{\left| {4x} \right|}}{3}+C$

Be careful! There are a lot of integrals that look very similar to the forms above but are actually different.

$\displaystyle \int{{\frac{{dx}}{{\sqrt{{{{x}^{2}}-1}}}}}}$

$\displaystyle \int{{\frac{{dx}}{{\sqrt{{1+{{x}^{2}}}}}}}}$

$\displaystyle \int{{\frac{{dx}}{{1-{{x}^{2}}}}}}$

Finding the integral of this forms: