Inverse Trigonometric Integration

Reference > Calculus: Integration

Description

\(\int arcsin(x) \, dx=xarcsin(x)+\sqrt{1-{x}^{2}}\)

\(\int arccos(x) \, dx=xarccos(x)-\sqrt{1-{x}^{2}}\)

\(\int arctan(x) \, dx=xarctan(x)-\frac{1}{2}\ln{(1+{x}^{2})}\)

\(\int arccsc(x) \, dx=xarccsc(x)-\ln{(x+x\sqrt{\frac{x-1}{{x}^{2}}})}\)

\(\int arcsec(x) \, dx=xarcsec(x)-\ln{(x+x\sqrt{\frac{x-1}{{x}^{2}}})}\)

\(\int arccot(x) \, dx=xarccot(x)+\frac{1}{2}\ln{(1+{x}^{2})}\)

Examples
\[\int arcsin(x) \, dx\]
1
Apply Integration By Parts to \(\int arcsin(x) \, dx\)
Let \(u=arcsin(x)\), \(dv=1\), \(du=\frac{1}{\sqrt{1-{x}^{2}}} dx\), \(v=x\)

2
Substitute the above into \(uv-\int v \, du\)
\[arcsin(x)x-\int \frac{x}{\sqrt{1-{x}^{2}}} \, dx\]

3
Apply Integration By Substitution to \(\int \frac{x}{\sqrt{1-{x}^{2}}} \, dx\)
Let \(u=1-{x}^{2}\), \(du=-2x dx\), then \(x dx=-\frac{1}{2} du\)

4
Using \(u\) and \(du\) above, rewrite \(\int \frac{x}{\sqrt{1-{x}^{2}}} \, dx\)
\[\int -\frac{1}{2\sqrt{u}} \, du\]

5
Apply the Constant Factor Rule: \(\int cf(x) \, dx=c\int f(x) \, dx\)
\[-\frac{1}{2}\int \frac{1}{\sqrt{u}} \, du\]

6
Since \(\frac{1}{\sqrt{x}} = {x}^{-\frac{1}{2}}\), using the Power Rule, \(\int {x}^{-\frac{1}{2}} \, dx=2{x}^{\frac{1}{2}}\)
\[-\sqrt{u}\]

7
Substitute \(u=1-{x}^{2}\) back into the original integral
\[-\sqrt{1-{x}^{2}}\]

8
Rewrite the integral with the completed substitution
\[arcsin(x)x+\sqrt{1-{x}^{2}}\]

9
Add constant
\[arcsin(x)x+\sqrt{1-{x}^{2}}+C\]

Done