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10.4 Usubstitution Trig Functionsap Calculus

The procedure for applying the Extreme Value Theorem is to first establish that the function is continuous on the closed interval. The next step is to determine all critical points in the given interval and evaluate the function at these critical points and at the endpoints of the interval. In calculus, trigonometric substitution is a technique for evaluating integrals. Moreover, one may use the trigonometric identities to simplify certain integrals containing radical expressions. 1 2 Like other methods of integration by substitution, when evaluating a definite integral, it may be simpler to completely deduce the.

Learn trig functions calculus ab trigonometric with free interactive flashcards. Choose from 500 different sets of trig functions calculus ab trigonometric flashcards on Quizlet. Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!!:)!! Calculus: U-substitution. Exponential and Natural Log Functions AP Calculus spends a great deal of time on exponential functions in the form of y = e x. Students should know that the value of e = 2.71828 In AP Calculus, we primarily use logs with base e, which are called natural logs (ln). Ln The inverse of a logarithmic function is an exponential function.

FINDING INTEGRALS USING THE METHOD OF TRIGONOMETRIC SUBSTITUTION

The following integration problems use the method of trigonometric (trig) substitution. It is a method for finding antiderivatives of functions which contain square roots of quadratic expressions or rational powers of the form $ displaystyle frac{n}{2}$ (where $n$ is an integer) of quadratic expressions. Examples of such expressions are $$ displaystyle{ sqrt{ 4-x^2 }} and displaystyle{(x^2+1)^{3/2}} $$The method of trig substitution may be called upon when other more common and easier-to-use methods of integration have failed. Trig substitution assumes that you are familiar with standard trigonometric identies, the use of differential notation, integration using u-substitution, and the integration of trigonometric functions. Recall that if$$ x = f(theta) , $$$$ dx = f'(theta) dtheta $$For example, if$$ x = sec theta , $$then$$ dx = sec theta tan theta dtheta $$The goal of trig substitution will be to replace square roots of quadratic expressions or rational powers of the form $ displaystyle frac{n}{2} $ (where $ n $ is an integer) of quadratic expressions, which may be impossible to integrate using other methods of integration, with integer powers of trig functions, which are more easily integrated. For example, if we start with the expression$$ displaystyle{ sqrt{4-x^2} } $$ and let$$ x = 2 sin theta , $$ then$$ displaystyle{ sqrt{4-x^2} } = displaystyle{ sqrt{4-(2 sin theta)^2 } } $$$$ = displaystyle{ sqrt{4-4 sin^2 theta } } $$$$ = displaystyle{ sqrt{4 (1- sin^2 theta ) } } $$$$ = displaystyle{ sqrt{4} cdot sqrt{1- sin^2 theta } } $$(Recall that $ cos^2 theta + sin^2 theta = 1 $ so that $ 1- sin^2 theta = cos^2 theta $.)$$ = displaystyle{2 cdot sqrt{ cos^2 theta } } $$ $$ = displaystyle{2 cdot Big cos theta Big } $$ (Assume that $ displaystyle - frac{pi}{2} le theta le displaystyle frac{pi}{2} $ so that $ cos theta ge 0 $.) $$ = displaystyle{2 cos theta } $$ and$$ dx = 2 cos theta d theta $$Thus,$$ displaystyle{ int sqrt{4-x^2} , dx } $$could be rewritten as$$ displaystyle{ int sqrt{4-x^2} , dx } = displaystyle{ int 2 cos theta cdot 2 cos theta , d theta } = displaystyle{ 4 int cos^2 theta , d theta } $$(Recall that $ cos 2 theta = 2 cos^2 theta -1 $ so that $ cos^2 theta = displaystyle frac{1}{2}(1+ cos 2 theta) $ .) $$ = displaystyle{ 4 int frac{1}{2}(1+ cos 2 theta) , d theta } $$$$ = displaystyle{ 2 int (1+ cos 2 theta) , d theta } $$$$ = displaystyle{ 2 ( theta + frac{1}{2} sin 2 theta) } + C $$ $$ = displaystyle{ 2 theta + sin 2 theta } + C $$(Recall that $ sin 2 theta = 2 sin theta cos theta $ .)$$ = displaystyle{ 2 theta + 2 sin theta cos theta } + C $$We need to write our final answer in terms of $ x$. Since $ x = 2 sin theta $, it follows that$$ sin theta = displaystyle{ x over 2} = displaystyle{ opposite over hypotenuse } $$ and $$ theta = arcsin Big(displaystyle frac{x}{2} Big) $$Using the given right triangle and the Pythagorean Theorem, we can determine any trig value of $ theta $.

Since $$ displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 longrightarrow $$ $$ (adjacent)^2 + (x)^2 = (2)^2 longrightarrow adjacent = sqrt{4-x^2} longrightarrow $$ $$ cos theta = displaystyle{ adjacent over hypotenuse }= displaystyle{ sqrt{4-x^2} over 2 } $$ Then $$ displaystyle{ 2 theta + 2 sin theta cos theta } + C = 2 arcsin Big( frac{x}{2} Big) + 2 cdot displaystyle{ x over 2} cdot displaystyle{ sqrt{4-x^2} over 2} $$$$ = displaystyle 2 arcsin Big( frac{x}{2} Big) + frac{1}{2} x cdot sqrt{4-x^2} + C $$ When using the method of trig substitution, we will always use one of the following three well-known trig identities :
10.4
  • (I) $ 1 - sin^2 theta = cos^2 theta $
  • (II) $ 1 + tan^2 theta = sec^2 theta $ and
  • (III) $ sec^2 theta - 1 = tan^2 theta $
For the expression $$ sqrt{a^2-x^2} $$we use equation (I) and let $$ x = a sin theta $$ (Assume that $ displaystyle - frac{pi}{2} le theta le displaystyle frac{pi}{2} $ so that $ cos theta ge 0 $. This allows for both positve and negative values of $ x$.)
Then $$ sqrt{a^2-x^2} = sqrt{a^2-a^2 sin^2 theta } $$$$ = sqrt{a^2(1- sin^2 theta)} $$$$ = sqrt{a^2 cos^2 theta} $$$$ = sqrt{a^2} sqrt{cos^2 theta} $$$$ = a sqrt{cos^2 theta} $$$$ = a big cos theta big $$$$ = a cos theta $$ and $$ dx = a cos theta d theta $$ For the expression $$ sqrt{a^2+x^2} $$we use equation (II) and let $$ x = a tan theta $$ (Assume that $ displaystyle - frac{pi}{2} le theta le displaystyle frac{pi}{2} $ so that $ cos theta > 0 $ and $ sec theta gt 0 $. This allows for both positve and negative values of $ x$.)

10.4 U-substitution Trig Functionsap Calculus Problems


Then $$ sqrt{a^2+x^2} = sqrt{a^2+a^2 tan^2 theta} $$$$ = sqrt{a^2(1+ tan^2 theta)} $$$$ = sqrt{a^2} sqrt{sec^2 theta} $$$$ = a sqrt{ sec^2 theta} $$$$ = a big sec theta big $$$$ = a sec theta $$and $$ dx = a sec^2 theta d theta $$ For the expression $$ sqrt{x^2-a^2} $$ we use equation (III) and let $$ x = a sec theta $$ (Assume that $ 0 le theta lt displaystyle frac{ pi }{2} , $ so that $ tan theta ge 0 $. This allows for only positive values of $ x$. If the integral includes negative values of $ x$, you must use $ displaystyle frac{pi}{2} lt theta le pi $ with $ sqrt{ tan^2 theta} = - tan theta $.)
Then $$ sqrt{x^2-a^2} = sqrt{a^2 sec^2 theta - a^2} $$$$ = sqrt{a^2(sec^2 theta - 1)} $$$$ = sqrt{a^2} sqrt{sec^2 theta - 1} $$$$ = a sqrt{ tan^2 theta} $$$$ = a big tan theta big $$$$ = a tan theta $$and $$ dx = a sec theta tan theta d theta $$ Recall the following well-known, basic indefinite trigonometric integral formulas :
  • 1.) $ displaystyle{ int cos x , dx } = sin x + C $
  • 2.) $ displaystyle{ int sin x , dx } = - cos x + C $
  • 3.) $ displaystyle{ int sec^2 x , dx } = tan x + C $
  • 4.) $ displaystyle{ int csc^2 x , dx } = - cot x + C $
  • 5.) $ displaystyle{ int sec x tan x , dx } = sec x + C $
  • 6.) $ displaystyle{ int csc x cot x , dx } = - csc x + C $
  • 7.) $ displaystyle{ int tan x , dx } = ln sec x + C $
  • 8.) $ displaystyle{ int cot x , dx } = ln sin x + C $
  • 9.) $ displaystyle{ int sec x , dx } = ln sec x + tan x + C $
  • 10.) $ displaystyle{ int csc x , dx } = ln csc x - cot x + C $

Most of the following problems are average. A few are somewhat challenging. Make careful and precise use of the differential notation $ dx $ and $ d theta $ and be careful when arithmetically and algebraically simplifying expressions. You should be proficient integrating various powers and rational functions involving trig functions. You may need to use the following additional well-known trig identities.
  • A.) $ sin 2x = 2 sin x cos x $
  • B.) $ cos 2x = 2 cos^2 x - 1 $ so that $ cos^2 x = displaystyle{ frac{1}{2}(1+cos 2x)} $
  • C.) $ cos 2x = 1 - 2 sin^2 x $ so that $ sin^2 x = displaystyle{ frac{1}{2} (1-cos 2x) } $
  • D.) $ cos 2x = cos^2 x - sin^2 x $
  • E.) $ 1 + cot^2 x = csc^2 x $ so that $ cot^2 x = csc^2 x - 1 $

  • PROBLEM 1 : Integrate $ displaystyle{ int { sqrt{1-x^2} } ,dx } $

    Click HERE to see a detailed solution to problem 1.

  • PROBLEM 2 : Integrate $ displaystyle{ int { (x^2-1)^{3/2} over x } , dx } $ .

    Click HERE to see a detailed solution to problem 2.

  • PROBLEM 3 : Integrate $ displaystyle{ int { 1 over (1-x^2)^{3/2} } , dx } $ .

    Click HERE to see a detailed solution to problem 3.

  • PROBLEM 4 : Integrate $ displaystyle{ int { sqrt{x^2+1} over x } , dx } $ .

    Activist learning projectmr.'s learning website. Click HERE to see a detailed solution to problem 4.

  • PROBLEM 5 : Integrate $ displaystyle{ int x^3 sqrt{4-9x^2} , dx } $ .

    Click HERE to see a detailed solution to problem 5.

  • PROBLEM 6 : Integrate $ displaystyle{ int { sqrt{1-x^2} over x } , dx } $ .

    Click HERE to see a detailed solution to problem 6.

  • PROBLEM 7 : Integrate $ displaystyle{ int { sqrt{x^2-9} over x^2 } , dx } $ .

    Click HERE to see a detailed solution to problem 7.

  • PROBLEM 8 : Integrate $ displaystyle{ int { sqrt{x^2+1} over x^2 } ,dx } $ .

    Click HERE to see a detailed solution to problem 8.

  • PROBLEM 9 : Integrate $ displaystyle{ int sqrt{x^2+25 } ,dx } $ .

    Click HERE to see a detailed solution to problem 9.

  • PROBLEM 10 : Integrate $ displaystyle{ int { sqrt{x^2-4} } ,dx } $ .

    Click HERE to see a detailed solution to problem 10.

  • PROBLEM 11 : Integrate $ displaystyle{ int { x over sqrt{x^4-16} } , dx } $ .

    Click HERE to see a detailed solution to problem 11.

  • PROBLEM 12 : Integrate $ displaystyle{ int { 1 over sqrt{x^2-4x} } , dx } $ .

    Click HERE to see a detailed solution to problem 12.

  • PROBLEM 13 : Integrate $ displaystyle{ int { x over sqrt{x^2+4x+5} } , dx } $ .

    Click HERE to see a detailed solution to problem 13.

  • PROBLEM 14 : Integrate $ displaystyle{ int { x cdot sqrt{10x-x^2} } , dx } $ .

    Click HERE to see a detailed solution to problem 14.

  • PROBLEM 15 : Integrate $ displaystyle{ int { sqrt{ {x-1} over x } } , dx } $ .

    Click HERE to see a detailed solution to problem 15.

  • PROBLEM 16 : Integrate $ displaystyle{ int { sqrt{1-x} cdot sqrt{x+3} } , dx } $ .

    Click HERE or HERE to see a detailed solution to problem 16.

  • PROBLEM 17 : A motorboat is resting at the origin, $(0, 0)$, and a skier tethered to the boat with a 20-foot long rope, is resting at the point $(20, 0)$. The boat then begins moving along the positive $y$-axis, pulling the skier along the unknown path $y=f(x)$. Use integration to find an equation for this path.

    Click HERE to see a detailed solution to problem 17.


    Click HERE to return to the original list of various types of calculus problems.

    Your comments and suggestions are welcome. Please e-mail any correspondence to Duane Kouba by clicking on the following address :

    A heartfelt 'Thank you' goes to The MathJax Consortium and the online Desmos Grapher for making the construction of graphs and this webpage fun and easy.
    Duane Kouba ..May 31, 2017

The following variables and constants are reserved:

  • e = Euler's number, the base of the exponential function (2.718281..)
  • i = imaginary number (i² = -1)
  • pi, π = the ratio of a circle's circumference to its diameter (3.14159..)
  • phi, Φ = the golden ratio (1,6180..)
  • t, u and v are used internally for integration by substitution and integration by parts

You can enter expressions the same way you see them in your math textbook. Implicit multiplication (5x = 5*x) is supported. If you are entering the integral from a mobile phone, you can also use ** instead of ^ for exponents. The interface is specifically optimized for mobile phones and small screens.

10.4 U-substitution Trig Functionsap Calculus Pdf

Supported integration rules and methods

The calculator decides which rule to apply and tries to solve the integral and find the antiderivative the same way a human would.

  • Integration of constants and constant functions
  • Integration by Subsitution (u-substitution)
  • Exponential and Logarithmic Functions
  • Trigonometric and Hyperbolic functions
  • Integration by splitting the function into partial fractions

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