# Derivatives and Integrals of Inverse Trig Functions

This section covers:

We learned about the Inverse Trig Functions here, and it turns out that the derivatives of them are not trig expressions, but algebraic. When memorizing these, remember that the functions starting with “$$c$$” are negative, and the functions with tan and cot don’t have a square root.
Also remember that sometimes you see the inverse trig function written as $$\arcsin x$$ and sometimes you see $${{\sin }^{{-1}}}x$$.

# Derivatives of Inverse Trig Functions

Here are the derivatives of Inverse Trigonometric Functions:

Derivatives of Inverse Trig Functions

$$\displaystyle \frac{{d\left( {\arcsin u} \right)}}{{dx}}=\frac{{{u}’}}{{\sqrt{{1-{{u}^{2}}}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{d\left( {\arccos u} \right)}}{{dx}}=\frac{{-{u}’}}{{\sqrt{{1-{{u}^{2}}}}}}$$

$$\displaystyle \frac{{d\left( {\arctan u} \right)}}{{dx}}=\frac{{{u}’}}{{1+{{u}^{2}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{d\left( {\text{arccot }u} \right)}}{{dx}}=\frac{{-{u}’}}{{1+{{u}^{2}}}}$$

$$\displaystyle \frac{{d\left( {\text{arcsec }u} \right)}}{{dx}}=\frac{{{u}’}}{{\left| u \right|\sqrt{{{{u}^{2}}-1}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{d\left( {\text{arccsc }u} \right)}}{{dx}}=\frac{{-{u}’}}{{\left| u \right|\sqrt{{{{u}^{2}}-1}}}}$$

Let’s try some problems:

 Inverse Trig Derivative Problem Solution Find the derivative: $$f\left( x \right)=\arcsin \left( {5x} \right)$$   ($$\displaystyle \frac{{dy}}{{dx}}\left( {\arcsin u} \right)=\frac{{{u}’}}{{\sqrt{{1-{{u}^{2}}}}}}$$) $$\displaystyle {f}’\left( x \right)=\frac{5}{{\sqrt{{1-{{{\left( {5x} \right)}}^{2}}}}}}=\frac{5}{{\sqrt{{1-25{{x}^{2}}}}}}$$ Find the derivative: $$\displaystyle y=\ln \left( {\frac{{2x+1}}{{2x-1}}} \right)+2\arctan \left( {2x} \right)$$   ($$\displaystyle \frac{{dy}}{{dx}}\left( {\arctan u} \right)=\frac{{{u}’}}{{1+{{u}^{2}}}}$$) \require {cancel} \displaystyle \begin{align}y&=\ln \left( {2x+1} \right)-\ln \left( {2x-1} \right)+2\arctan \left( {2x} \right)\\{y}’&=\frac{2}{{2x+1}}-\frac{2}{{2x-1}}+\frac{{2\cdot 2}}{{1+{{{\left( {2x} \right)}}^{2}}}}=\frac{2}{{2x+1}}-\frac{2}{{2x-1}}+\frac{4}{{4{{x}^{2}}+1}}\\&=\frac{{2\left( {2x-1} \right)\left( {4{{x}^{2}}+1} \right)}}{{\left( {4{{x}^{2}}-1} \right)\left( {4{{x}^{2}}+1} \right)}}-\frac{{2\left( {2x+1} \right)\left( {4{{x}^{2}}+1} \right)}}{{\left( {4{{x}^{2}}-1} \right)\left( {4{{x}^{2}}+1} \right)}}+\frac{{4\left( {4{{x}^{2}}-1} \right)}}{{\left( {4{{x}^{2}}-1} \right)\left( {4{{x}^{2}}+1} \right)}}\\&=\frac{{\cancel{{16{{x}^{3}}}}\cancel{{-8{{x}^{2}}}}+\cancel{{4x}}-2-\left( {\cancel{{16{{x}^{3}}}}+\cancel{{8{{x}^{2}}}}+\cancel{{4x}}+2} \right)+\cancel{{16{{x}^{2}}}}-4}}{{\left( {4{{x}^{2}}-1} \right)\left( {4{{x}^{2}}+1} \right)}}\\&=\frac{{-8}}{{16{{x}^{4}}-1}}=\frac{8}{{1-16{{x}^{4}}}}\end{align} (May not have to simplify this much!) Find the derivative: $$\displaystyle y=\frac{{\arccos \left( {5x} \right)}}{{{{x}^{2}}}}$$   ($$\displaystyle \frac{{dy}}{{dx}}\left( {\arccos u} \right)=\frac{{-{u}’}}{{\sqrt{{1-{{u}^{2}}}}}}$$) \begin{align}{y}’&=\frac{{{{x}^{2}}\left( {\frac{{-5}}{{\sqrt{{1-{{{\left( {5x} \right)}}^{2}}}}}}} \right)-\arccos \left( {5x} \right)\cdot 2x}}{{{{{\left( {{{x}^{2}}} \right)}}^{2}}}}=\frac{{\frac{{-5{{x}^{2}}}}{{\sqrt{{1-25{{x}^{2}}}}}}-2x\cdot \arccos \left( {5x} \right)}}{{{{x}^{4}}}}\\&=\frac{{-5{{x}^{2}}}}{{{{x}^{4}}\sqrt{{1-25{{x}^{2}}}}}}-\frac{{2x\cdot \arccos \left( {5x} \right)}}{{{{x}^{4}}}}=\frac{{-5}}{{{{x}^{2}}\sqrt{{1-25{{x}^{2}}}}}}-\frac{{2\arccos \left( {5x} \right)}}{{{{x}^{3}}}}\end{align} Find an equation of tangent line of the function at the given point: $$\displaystyle f\left( x \right)=\text{arcsec}\sqrt{x}\,,\,\,\,\,\,\left( {\frac{4}{3},\,\,\frac{\pi }{6}} \right)$$   ($$\displaystyle \frac{{dy}}{{dx}}\left( {\text{arcsec}\,u} \right)=\frac{{{u}’}}{{\left| u \right|\sqrt{{{{u}^{2}}-1}}}}$$) $$\displaystyle f\left( x \right)=\text{arcsec}\left( {{{x}^{{\frac{1}{2}}}}} \right);\,\,\,\,{f}’\left( x \right)=\frac{{\frac{1}{2}{{x}^{{-\frac{1}{2}}}}}}{{\left| {\sqrt{x}} \right|\sqrt{{\left( {{{{\left( {\sqrt{x}} \right)}}^{2}}} \right)-1}}}}=\frac{1}{{2\left| {\sqrt{x}} \right|\sqrt{{x-1}}\sqrt{x}}}=\frac{1}{{2x\sqrt{{x-1}}}}$$ At $$\displaystyle \,\left( {\frac{4}{3},\frac{\pi }{6}} \right)$$, $$\displaystyle {f}’\left( {\frac{4}{3}} \right)=\frac{1}{{2\left( {\frac{4}{3}} \right)\sqrt{{\frac{4}{3}-1}}}}=\frac{3}{{8\sqrt{{\frac{1}{3}}}}}=\frac{{3\sqrt{3}}}{8}$$ Tangent line: $$\displaystyle y-\frac{\pi }{6}=\frac{{3\sqrt{3}}}{8}\left( {x-\frac{4}{3}} \right);\,\,\,\,y=\frac{{3\sqrt{3}}}{8}x-\frac{{\sqrt{3}}}{2}+\frac{\pi }{6}$$ Use implicit differentiation to find $${y}’$$: $$\text{arccsc}\,x+\text{arccot}\left( {x+y} \right)=\pi$$   ($$\displaystyle \frac{{dy}}{{dx}}\left( {\text{arccsc}\,u} \right)=\frac{{-{u}’}}{{\left| u \right|\sqrt{{{{u}^{2}}-1}}}}$$; $$\displaystyle \frac{{dy}}{{dx}}\left( {\text{arccot}\,u} \right)=\frac{{-{u}’}}{{1+{{u}^{2}}}}$$) $$\displaystyle \text{arccsc}\,x+\text{arccot}\left( {x+y} \right)=\pi ;\,\,\,\,\,\,\,\frac{{-1}}{{\left| x \right|\sqrt{{{{x}^{2}}-1}}}}+\frac{{-\left( {1+{y}’} \right)}}{{1+{{{\left( {x+y} \right)}}^{2}}}}=0$$ $$\displaystyle \frac{{\left( {1+{y}’} \right)}}{{1+{{{\left( {x+y} \right)}}^{2}}}}=\frac{{-1}}{{\left| x \right|\sqrt{{{{x}^{2}}-1}}}};\,\,\,\,\,\,\left( {1+{y}’} \right)=\frac{{-1}}{{\left| x \right|\sqrt{{{{x}^{2}}-1}}}}\left[ {1+{{{\left( {x+y} \right)}}^{2}}} \right]$$ $$\displaystyle {y}’=\frac{{-1-{{{\left( {x+y} \right)}}^{2}}}}{{\left| x \right|\sqrt{{{{x}^{2}}-1}}}}-1$$

# Integrals Involving the Inverse Trig Functions

When we integrate to get Inverse Trigonometric Functions back, we have use tricks to get the functions to look like one of the inverse trig forms and then usually use U-Substitution Integration to perform the integral.

Here are the integration formulas involving the Inverse Trig Functions; notice that we only have formulas for three of the inverse trig functions; trust me, it works this way!

To the right of each formula, I’ve included a short-cut formula that you may want to learn; however, if you just know the first formulas at the left (that resemble the differentiation formulas), you will be able to use U-substitution to solve the problems.

Integrals Involving the Inverse Trig Functions

\displaystyle \begin{align}\int{{\frac{{du}}{{\sqrt{{1-{{u}^{2}}}}}}}}\,=\arcsin \,u+C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int{{\frac{{du}}{{\sqrt{{{{a}^{2}}-{{u}^{2}}}}}}}}\,=\arcsin \,\frac{u}{a}+C\\\int{{\frac{{du}}{{1+{{u}^{2}}}}}}\,=\arctan \,u+C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int{{\frac{{du}}{{{{a}^{2}}+{{u}^{2}}}}}}\,=\frac{1}{a}\arctan \,\frac{u}{a}+C\\\int{{\frac{{du}}{{u\sqrt{{{{u}^{2}}-1}}}}}}\,=\text{arcsec}\,\left| u \right|+C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int{{\frac{{du}}{{u\sqrt{{{{u}^{2}}-{{a}^{2}}}}}}}}\,=\frac{1}{a}\text{arcsec}\,\frac{{\left| u \right|}}{a}+C\end{align}

A lot of times, to get the integral in the correct form, we have to play with the function to get a “1” in the denominator, either in the square root, or without it (for tan and cot). To do this, just take the greatest common factor (GCF) of the constant out, so a “1” will remain; we’ll see this in problems below. Sometimes, we’ll also have to Complete the Square, as shown below.

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Let’s first do some Indefinite Integration problems:

 Integration to get Inverse Trig Function Solution Find the indefinite integral:   $$\displaystyle \int{{\frac{{dx}}{{\sqrt{{1-25{{x}^{2}}}}}}}}$$ \displaystyle \begin{align}u&=5x\\du&=5dx\\dx&=\frac{{du}}{5}\end{align}                  \begin{align}\int{{\frac{{dx}}{{\sqrt{{1-25{{x}^{2}}}}}}}}&=\int{{\frac{{dx}}{{\sqrt{{1-{{{\left( {5x} \right)}}^{2}}}}}}=}}\int{{\frac{1}{{\sqrt{{1-{{u}^{2}}}}}}\cdot \frac{{du}}{5}}}\\&=\frac{1}{5}\int{{\frac{{du}}{{\sqrt{{1-{{u}^{2}}}}}}=}}\frac{1}{5}\arcsin u+C=\frac{1}{5}\arcsin \left( {5x} \right)+C\end{align} Find the indefinite integral:   $$\displaystyle \int{{\frac{{dx}}{{x\sqrt{{16{{x}^{2}}-4}}}}}}$$ \displaystyle \begin{align}u&=2x\\du&=2dx\\dx&=\frac{{du}}{2}\\x&=\frac{u}{2}\end{align}      \require {cancel} \displaystyle \begin{align}\int{{\frac{{dx}}{{x\sqrt{{16{{x}^{2}}-4}}}}}}&=\int{{\frac{{dx}}{{x\sqrt{{4\left[ {{{{\left( {2x} \right)}}^{2}}-1} \right]}}}}=}}\int{{\frac{{dx}}{{2x\sqrt{{{{{\left( {2x} \right)}}^{2}}-1}}}}=}}\frac{1}{2}\int{{\frac{{\frac{{du}}{{\cancel{2}}}}}{{\frac{u}{{\cancel{2}}}\sqrt{{{{u}^{2}}-1}}}}}}\\&=\frac{1}{2}\int{{\frac{{du}}{{u\sqrt{{{{u}^{2}}-1}}}}}}=\frac{1}{2}\text{arcsec}\left| u \right|+C=\frac{1}{2}\text{arcsec}\left| {2x} \right|+C\end{align} Find the indefinite integral:   $$\displaystyle \int{{\frac{t}{{{{t}^{4}}+9}}}}\,dt$$ \begin{align}u&=\frac{{{{t}^{2}}}}{3}\\du&=\frac{{2t}}{3}dt\,\\dt&=\frac{3}{{2t}}du\end{align}       \displaystyle \begin{align}\int{{\frac{t}{{{{t}^{4}}+9}}}}\,dt&=\int{{\frac{t}{{9\left( {{{{\left( {\frac{{{{t}^{2}}}}{3}} \right)}}^{2}}+1} \right)}}}}\,dt=\frac{1}{{{}_{3}\cancel{9}}}\int{{\frac{{\cancel{t}}}{{{{u}^{2}}+1}}}}\cdot \frac{{\cancel{3}}}{{2\cancel{t}}}du=\frac{1}{6}\int{{\frac{t}{{{{u}^{2}}+1}}}}\,du\\&=\frac{1}{6}\arctan u+C=\frac{1}{6}\arctan \left( {\frac{{{{t}^{2}}}}{3}} \right)+C\end{align} Find the indefinite integral:   $$\displaystyle \int{{\frac{{{{{\sec }}^{2}}x}}{{\sqrt{{36-{{{\tan }}^{2}}x}}}}}}dx$$ \displaystyle \begin{align}u&=\frac{{\tan x}}{6}\\du&=\frac{{{{{\sec }}^{2}}x}}{6}dx\\dx&=\frac{{6\,du}}{{{{{\sec }}^{2}}x}}\end{align}       \displaystyle \begin{align}\int{{\frac{{{{{\sec }}^{2}}x}}{{\sqrt{{36-{{{\tan }}^{2}}x}}}}}}\,dx&=\int{{\frac{{{{{\sec }}^{2}}x}}{{\sqrt{{36\left[ {1-{{{\left( {\frac{{\tan x}}{6}} \right)}}^{2}}} \right]}}}}\,}}dx=\frac{1}{{\cancel{6}}}\int{{\frac{{\cancel{{{{{\sec }}^{2}}x}}}}{{\sqrt{{1-{{u}^{2}}}}}}}}\cdot \frac{{\cancel{6}\,du}}{{\cancel{{{{{\sec }}^{2}}x}}}}\\&=\arcsin u+C=\arcsin \left( {\frac{{\tan x}}{6}} \right)+C\end{align}

Here are a few more integrals involving inverse trig functions that are bit more complicated. Note that we need to Complete the Square in the last problem.

 Integration to get Inverse Trig Function Solution Find the indefinite integral:   $$\displaystyle \int{{\frac{{x+8}}{{\sqrt{{16-{{{\left( {x-4} \right)}}^{2}}}}}}}}\,dx$$   (Trick: Separate the integral into two terms that we can integrate.  The first numerator ($$x-4$$ , in this case) should match what is squared in the denominator). \displaystyle \begin{align}&\text{1st term: }\\u&=16-{{\left( {x-4} \right)}^{2}}\\du&=\left( {-2x+8} \right)dx\\dx&=\frac{{-du}}{{2x-8}}\\\\\\&\text{ 2nd term: }\\v&=\frac{{x-4\text{ }}}{4}\\dv&=\frac{1}{4}dx\\dx&=4dv\\\end{align}          \require {cancel} \displaystyle \begin{align}&\,\,\,\,\,\,\,\,\int{{\frac{{x+8}}{{\sqrt{{16-{{{\left( {x-4} \right)}}^{2}}}}}}}}\,dx\\&=\int{{\frac{{x-4}}{{\sqrt{{16-{{{\left( {x-4} \right)}}^{2}}}}}}}}\,dx+\int{{\frac{{12}}{{\sqrt{{16-{{{\left( {x-4} \right)}}^{2}}}}}}}}\,dx\\&=\int{{\left( {x-4} \right){{{\left[ {16-{{{\left( {x-4} \right)}}^{2}}} \right]}}^{{-\frac{1}{2}}}}}}dx+\int{{\frac{{12}}{{4\sqrt{{1-{{{\left( {\frac{{x-4}}{4}} \right)}}^{2}}}}}}}}\,dx\\&=\int{{\cancel{{\left( {x-4} \right)}}{{u}^{{-\frac{1}{2}}}}}}\cdot \frac{{-\,du}}{{{{{\cancel{{2x-8}}}}_{2}}}}+12\int{{\frac{{\cancel{4}\,dv}}{{\cancel{4}\sqrt{{1-{{v}^{2}}}}}}}}\\&=-\frac{1}{2}\int{{{{u}^{{-\frac{1}{2}}}}}}du+12\int{{\frac{{dv}}{{\sqrt{{1-{{v}^{2}}}}}}}}=-{{u}^{{\frac{1}{2}}}}+12\arcsin v+C\\&=-\sqrt{{16-{{{\left( {x-4} \right)}}^{2}}}}+12\arcsin \left( {\frac{{x-4}}{4}} \right)+C\end{align} Find the indefinite integral:   $$\displaystyle \int{{\frac{3}{{x\sqrt{{{{x}^{4}}-9}}}}dx}}$$ \displaystyle \begin{align}u&=\frac{{{{x}^{2}}}}{3}\\du&=\frac{{2x}}{3}dx\\dx&=\frac{{3\,}}{{2x}}du\end{align}                   \displaystyle \begin{align}&\,\,\,\,\,\int{{\frac{3}{{x\sqrt{{{{x}^{4}}-9}}}}dx}}\\&=\int{{\frac{3}{{x\sqrt{{\,9\left( {\frac{{{{x}^{4}}}}{9}-1} \right)}}}}dx}}=\int{{\frac{{\cancel{3}}}{{\cancel{3}x\sqrt{{\frac{{{{x}^{4}}}}{9}-1}}}}dx}}\\&=\int{{\frac{1}{{x\sqrt{{{{{\left( {\frac{{{{x}^{2}}}}{3}} \right)}}^{2}}-1}}}}dx}}=\int{{\frac{1}{{x\sqrt{{{{u}^{2}}-1}}}}\cdot \left( {\frac{3}{{2x}}} \right)du}}\end{align} This looks like an arcsec integral ($$\displaystyle \int{{\frac{{du}}{{u\sqrt{{{{u}^{2}}-1}}}}}}$$), but it doesn’t quite fit (we have extra $$x$$’s on the bottom instead of a “$$u$$”).  So, what can we do? Method 1:  Since we need a $$\displaystyle \frac{{{{x}^{2}}}}{3}$$ for the “$$u$$”, multiply the numerator and denominator by $$\displaystyle \frac{x}{3}$$ to get it: \displaystyle \begin{align}\int{{\frac{1}{{x\sqrt{{{{u}^{2}}-1}}}}\cdot \left( {\frac{3}{{2x}}} \right)du}}&=\int{{\frac{{1\cdot \frac{x}{3}}}{{x\cdot \frac{x}{3}\sqrt{{{{u}^{2}}-1}}}}\cdot \left( {\frac{3}{{2x}}} \right)du}}=\int{{\frac{{\frac{{\cancel{x}}}{{\cancel{3}}}}}{{u\sqrt{{{{u}^{2}}-1}}}}\cdot \left( {\frac{{\cancel{3}}}{{2\cancel{x}}}} \right)du}}\\&=\frac{1}{2}\int{{\frac{1}{{u\sqrt{{{{u}^{2}}-1}}}}du}}=\frac{1}{2}\text{arcsec} \left| {\frac{{{{x}^{2}}}}{3}} \right|=\frac{1}{2}\text{arcsec} \left( {\frac{{{{x}^{2}}}}{3}} \right)+C\end{align} Method 2:  Solve for $$x$$ in terms of $$u$$ and plug in; it works: $$\displaystyle u=\frac{{{{x}^{2}}}}{3};\,\,\,\,\,{{x}^{2}}=3u;\,\,\,\,\,x=\pm \sqrt{{3u}}$$: \displaystyle \begin{align}&\,\,\,\,\,\,\,\int{{\frac{1}{{x\sqrt{{{{u}^{2}}-1}}}}\cdot \left( {\frac{3}{{2x}}} \right)du}}\\&=\int{{\frac{1}{{\left( {\pm \sqrt{{3u}}} \right)\sqrt{{{{u}^{2}}-1}}}}\cdot \left( {\frac{3}{{2\left( {\pm \sqrt{{3u}}} \right)}}} \right)du}}=\int{{\frac{3}{{6u\sqrt{{{{u}^{2}}-1}}}}du}}=\frac{1}{2}\text{arcsec}\left( {\frac{{{{x}^{2}}}}{3}} \right)+C\end{align} Find the indefinite integral:   $$\displaystyle \int{{\frac{1}{{{{x}^{2}}-4x+5}}dx}}$$ \begin{align}u&=x-2\\du&=\,dx\end{align}                 \displaystyle \begin{align}\int\limits_{{}}^{{}}{{\frac{1}{{{{x}^{2}}-4x+5}}}}\,dx&=\int\limits_{{}}^{{}}{{\frac{1}{{\left( {{{x}^{2}}-4x+4} \right)+5-4}}}}\,dx\\&=\int\limits_{{}}^{{}}{{\frac{1}{{{{{\left( {x-2} \right)}}^{2}}+1}}dx}}=\int\limits_{{}}^{{}}{{\frac{{\,1}}{{{{u}^{2}}+1}}}}du\\&=\arctan u+C=\arctan \left( {x-2} \right)+C\end{align}

Now let’s do some Inverse Trig Definite Integration Problems. Notice in the second problem, we have to use “$$\arcsin x$$” for “$$u$$”, and we need to Complete the Square for the last problem.

 Inverse Trig Integration Solution Evaluate the definite integral:   $$\displaystyle \int\limits_{0}^{{\ln 4}}{{\frac{{{{e}^{{-t}}}}}{{-1-{{e}^{{-2t}}}}}}}\,dt$$ \begin{align}{l}u&={{e}^{{-t}}}\\du&=-{{e}^{{-t}}}dt\\dt&=-\frac{{du}}{{{{e}^{{-t}}}}}\end{align}       \displaystyle \begin{align}\int\limits_{0}^{{\ln 4}}{{\frac{{{{e}^{{-t}}}}}{{-1-{{e}^{{-2t}}}}}}}\,dt&=\int\limits_{0}^{{\ln 4}}{{\frac{{{{e}^{{-t}}}}}{{-1-{{{\left( {{{e}^{{-t}}}} \right)}}^{2}}}}}}\,dt=\int\limits_{{t=0}}^{{t=\ln 4}}{{\frac{{\cancel{{{{e}^{{-t}}}}}}}{{-1-{{u}^{2}}}}}}\,\cdot \frac{{-du}}{{\cancel{{{{e}^{{-t}}}}}}}=\int\limits_{{t=0}}^{{t=\ln 4}}{{\frac{{du}}{{1+{{u}^{2}}}}}}\\\,\,\,&=\left[ {\arctan u} \right]_{{t=0}}^{{t=\ln 4}}=\left[ {\arctan {{e}^{{-t}}}} \right]_{0}^{{\ln 4}}=\arctan \left( {{{e}^{{-\ln 4}}}} \right)-\arctan \left( {{{e}^{0}}} \right)\\\,\,\,\,&=\arctan \left( {{{e}^{{\ln \left( {\frac{1}{4}} \right)}}}} \right)-\frac{\pi }{4}=\arctan \left( {\frac{1}{4}} \right)-\frac{\pi }{4}\approx -.54\end{align} Evaluate the definite integral:   $$\displaystyle \int\limits_{0}^{{\frac{{\sqrt{3}}}{2}}}{{\frac{{\arcsin x}}{{\sqrt{{1-{{x}^{2}}}}}}}}\,dx$$ \begin{align}u&=\arcsin x\\du&=\frac{1}{{\sqrt{{1-{{x}^{2}}}}}}\,dx\\dx&=\sqrt{{1-{{x}^{2}}}}\,du\\\end{align}                 \begin{align}\int\limits_{0}^{{\frac{{\sqrt{3}}}{2}}}{{\frac{{\arcsin x}}{{\sqrt{{1-{{x}^{2}}}}}}}}\,dx&=\int\limits_{{x=0}}^{{x=\frac{{\sqrt{3}}}{2}}}{{\frac{u}{{\cancel{{\sqrt{{1-{{x}^{2}}}}}}}}}}\cdot \,\cancel{{\sqrt{{1-{{x}^{2}}}}}}du=\left[ {\frac{1}{2}{{u}^{2}}} \right]_{{x=0}}^{{x=\frac{{\sqrt{3}}}{2}}}\\\,\,\,&=\left[ {\frac{1}{2}{{{\arcsin }}^{2}}x} \right]_{0}^{{\frac{{\sqrt{3}}}{2}}}=\frac{1}{2}{{\left( {\arcsin \frac{{\sqrt{3}}}{2}} \right)}^{2}}-\frac{1}{2}{{\left( {\arcsin 0} \right)}^{2}}\\\,&=\frac{1}{2}{{\left( {\frac{\pi }{3}} \right)}^{2}}-\frac{1}{2}\cdot 0=\frac{{{{\pi }^{2}}}}{{18}}\approx .548\end{align} Evaluate the definite integral:   $$\displaystyle \int\limits_{{-1}}^{1}{{\frac{1}{{{{x}^{2}}-4x+8}}}}\,dx$$ \begin{align}u&=\frac{{x-2}}{2}\\du&=\frac{1}{2}\,dx\\dx&=2\,du\end{align}        \displaystyle \begin{align}\int\limits_{{-1}}^{1}{{\frac{1}{{{{x}^{2}}-4x+8}}}}\,dx&=\int\limits_{{-1}}^{1}{{\frac{1}{{\left( {{{x}^{2}}-4x+4} \right)-4+8}}}}\,dx=\int\limits_{{-1}}^{1}{{\frac{{dx}}{{{{{\left( {x-2} \right)}}^{2}}+4}}}}\\\,\,\,&=\int\limits_{{-1}}^{1}{{\frac{{dx}}{{4\left[ {{{{\left( {\frac{{x-2}}{2}} \right)}}^{2}}+1} \right]}}}}=\frac{1}{4}\int\limits_{{-1}}^{1}{{\frac{{dx}}{{{{{\left( {\frac{{x-2}}{2}} \right)}}^{2}}+1}}}}=\frac{1}{4}\int\limits_{{x=-1}}^{{x=1}}{{\frac{{2\,du}}{{{{u}^{2}}+1}}}}\\\,\,&=\left[ {\frac{1}{2}\arctan \,u} \right]_{{x=-1}}^{{x=1}}=\left[ {\frac{1}{2}\arctan \,\frac{{x-2}}{2}} \right]_{{-1}}^{1}\\\,\,\,\,\,\,\,\,\,\,&=\frac{1}{2}\arctan \left( {-\frac{1}{2}} \right)-\frac{1}{2}\arctan \left( {-\frac{3}{2}} \right)\approx .2596\end{align}

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On to Applications of Integration: Area and Volume – you are ready!

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