Derivatives and Integrals of Inverse Trig Functions

We learned about the Inverse Trig Functions here, and it turns out that the derivatives of them are not trigonometric 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}}}}$

Here are 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}}}}$)

($ \displaystyle \frac{d}{{dx}}\left( {\ln u} \right)=\frac{{{u}’}}{u}$)

$ \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}}}}}}$)

(Use Quotient Rule)

$ \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( {\sqrt{x}} \right)}}^{2}}-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 \frac{{-1}}{{\left| x \right|\sqrt{{{{x}^{2}}-1}}}}+\frac{{-\left( {1+{y}’} \right)}}{{1+{{{\left( {x+y} \right)}}^{2}}}}=0$

$ \displaystyle \begin{align}\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]\\{y}’&=\frac{{-1-{{{\left( {x+y} \right)}}^{2}}}}{{\left| x \right|\sqrt{{{{x}^{2}}-1}}}}-1=\frac{{-1-{{{\left( {x+y} \right)}}^{2}}}}{{\sqrt{{{{x}^{2}}}}\sqrt{{{{x}^{2}}-1}}}}-\frac{{\sqrt{{{{x}^{2}}}}\sqrt{{{{x}^{2}}-1}}}}{{\sqrt{{{{x}^{2}}}}\sqrt{{{{x}^{2}}-1}}}}\\&=\frac{{-1-{{x}^{2}}-2xy-{{y}^{2}}-\sqrt{{{{x}^{2}}}}\sqrt{{{{x}^{2}}-1}}}}{{\sqrt{{{{x}^{2}}}}\sqrt{{{{x}^{2}}-1}}}}\end{align}$

(There are many ways to “simplify”; I used $ \left| x \right|=\sqrt{{{{x}^{2}}}}$ to get rid of the the absolute value sign.)

Integrals Involving Inverse Trig Functions (Integrals Resulting in Inverse Trigonometric 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; notice that we only have formulas for three of the inverse trig functions; trust me, it works this way (although there may be other correct answers)! 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-sub 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$” in the denominator, either in the square root, or without it (for tan and cot). To do this, take the greatest common factor (GCF) of the constant out, so a “$ 1$” will remain; we’ll see this in problems below. Sometimes, we’ll also have to Complete the Square, as shown below.

Note that we don’t really know what the “$ \boldsymbol {u}$” is until we factor out the GCF – until after we’ve “made it fit” into one of the formulas above!

Here are some of the more straight-forward Indefinite Integration problems; these can be tricky!

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\\x&=\frac{u}{2}\\du&=2dx\\dx&=\frac{{du}}{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{1}{{{{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}{{{{{\sec }}^{2}}x}}du\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}\,}}{{\cancel{{{{{\sec }}^{2}}x}}}}du\\&=\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}}$

 

(Remember that “in real life”, these integrals probably won’t fit into the mold that these problems do; somebody made these problems up to show you these neat tricks! They are still neat tricks though!)

$ \displaystyle \begin{align}u&=\frac{{{{x}^{2}}}}{3}\\(x&=\pm \sqrt{{3u}})\\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 \frac{3}{{2x}}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$”). What can we do?

Method 1:  Since we need $ \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 \frac{3}{{2x}}du}}&=\int{{\frac{{1\cdot \frac{x}{3}}}{{x\cdot \frac{x}{3}\sqrt{{{{u}^{2}}-1}}}}\cdot \frac{3}{{2x}}du}}=\int{{\frac{{\frac{{\cancel{x}}}{{\cancel{3}}}}}{{u\sqrt{{{{u}^{2}}-1}}}}\cdot \frac{{\cancel{3}}}{{2\cancel{x}}}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 \int{{\frac{1}{{x\sqrt{{{{u}^{2}}-1}}}}\cdot \frac{3}{{2x}}du}}=\int{{\frac{1}{{\left( {\pm \sqrt{{3u}}} \right)\sqrt{{{{u}^{2}}-1}}}}\cdot \frac{3}{{2\left( {\pm \sqrt{{3u}}} \right)}}du}}=\int{{\frac{3}{{6u\sqrt{{{{u}^{2}}-1}}}}du}}=\frac{1}{2}\text{arcsec}\left( {\frac{{{{x}^{2}}}}{3}} \right)+C$

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$

$ \displaystyle \begin{align}u&={{e}^{{-t}}}\\du&=-{{e}^{{-t}}}dt\\dt&=-\frac{1}{{{{e}^{{-t}}}}}du=-{{e}^{t}}du\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 -\cancel{{{{e}^{t}}}}du=\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 -.540\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}$

Learn these rules, and practice, practice, practice!


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