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Indefinite Integrals of Trig FunctionsIndefinite Integrals of Trig Functions
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Question 1 of 6
1. Question
Find the integral`int sin8x dx`Hint
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Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`Integrating Trigonometric Functions
$$\int f(\color{#004ec4}{g(x)}) dx=f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$Substitute the components into the formula$$\int f(\color{#004ec4}{g(x)}) dx$$ `=` $$f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$ $$\int \text{sin}(\color{#004ec4}{8x}) dx$$ `=` $$-\text{cos}\;8x\cdot\frac{1}{\color{#004ec4}{g'(8x)}} +c$$ Substitute known values `=` $$-\text{cos}\;8x\cdot\frac{1}{8} +c$$ Evaluate `=` $$-\frac{1}{8}\;\text{cos}\;8x +c$$ `-1/8 \text(cos) 8x+c` -
Question 2 of 6
2. Question
Find the integral`int 4sec^2 2x dx`Hint
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Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`Integrating Trigonometric Functions
$$\int f(\color{#004ec4}{g(x)}) dx=f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$Substitute the components into the formula$$\int f(\color{#004ec4}{g(x)}) dx$$ `=` $$f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$ $$\int \text{sec}^2(\color{#004ec4}{2x}) dx$$ `=` $$\text{tan}\;2x\cdot\frac{1}{\color{#004ec4}{g'(2x)}} +c$$ Substitute known values `=` $$\text{tan}\;2x\cdot\frac{1}{2} +c$$ Evaluate `=` $$2\;\text{tan}\;2x +c$$ `2 \text(tan) 2x+c` -
Question 3 of 6
3. Question
Find the integral`int sin(pi-x) dx`Hint
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Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`Integrating Trigonometric Functions
$$\int f(\color{#004ec4}{g(x)}) dx=f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$Substitute the components into the formula$$\int f(\color{#004ec4}{g(x)}) dx$$ `=` $$f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$ $$\int \text{sin}(\color{#004ec4}{\pi-x}) dx$$ `=` $$-\text{cos}\;(\pi-x)\cdot\frac{1}{\color{#004ec4}{g'(\pi-x)}} +c$$ Substitute known values `=` $$-\text{cos}\;(\pi-x)\cdot\frac{1}{-1} +c$$ Evaluate `=` $$\text{cos}\;(\pi-x) +c$$ `\text(cos) (\pi-x)+c` -
Question 4 of 6
4. Question
Find the integral`int [sec^2 2x-cos x/2] dx`Hint
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Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`Integrating Trigonometric Functions
$$\int f(\color{#004ec4}{g(x)}) dx=f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$Substitute the components of each term into the formulaFirst term$$\int f(\color{#004ec4}{g(x)}) dx$$ `=` $$f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$ $$\int \text{sec}^2\color{#004ec4}{2x}\;dx$$ `=` $$\text{tan}\;2x\cdot\frac{1}{\color{#004ec4}{g'(2x)}}$$ Substitute known values `=` $$\text{tan}\;2x\cdot\frac{1}{2}$$ Evaluate `=` $$\frac{1}{2}\text{tan}\;2x$$ Second term$$\int f(\color{#004ec4}{g(x)}) dx$$ `=` $$f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$ $$\int -\text{cos}\left(\color{#004ec4}{\frac{x}{2}}\right)\;dx$$ `=` $$-\text{sin}\;\frac{x}{2}\cdot\frac{1}{\color{#004ec4}{g'(\frac{x}{2})}}$$ Substitute known values `=` $$-\text{cos}\;\frac{x}{2}\cdot2$$ Evaluate `=` $$-2\text{cos}\;\frac{x}{2}$$ Finally, combine the two terms and add the constant$$\frac{1}{2}\text{tan}\;2x-2\text{sin}\;\frac{x}{2} +c$$ `1/2 \text(tan) 2x-2 \text(sin) x/2+c` -
Question 5 of 6
5. Question
Find the integral`int tan^2 x dx`Hint
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Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`Integrating Trigonometric Functions
$$\int f(\color{#004ec4}{g(x)}) dx=f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$First, convert the function into a derivable functionTake note that `\text(sec)^2=1+\text(tan)^2x``\text(sec)^2` `=` `1+\text(tan)^2x` `\text(sec)^2` `-1` `=` `1+\text(tan)^2x` `-1` Subtract `1` from both sides `\text(sec)^2-1` `=` `\text(tan)^2x` Therefore, we can use `\text(sec)^2-1` as a derivable substituteFinally, substitute the components into the formula$$\int f(\color{#004ec4}{g(x)}) dx$$ `=` $$f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$ $$\int \text{sec}^2\color{#004ec4}{x-1}\;dx$$ `=` $$\text{tan}\;x-x +c$$ Substitute known values and integrate `\text(tan) x-x+c` -
Question 6 of 6
6. Question
Find the integral`int tan x dx`Hint
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Integrals of Trigonometric Functions
`int \text(cos)=\text(sin)``int \text(sin)=-\text(cos)``int \text(sec)^2=\text(tan)`Integrating Trigonometric Functions
$$\int f(\color{#004ec4}{g(x)}) dx=f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$$First, convert the function into a derivable functionTake note that `\text(tan) x=(\text(sin) x)/(\text(cos) x)`Therefore, we can use `(\text(sin) x)/(\text(cos) x)` as a derivable substituteNext, take note that `d/(dx)\text(cos) x=-\text(sin) x`This means that the function satisfies the derivative of a natural logarithm `(f'(x))/f(x)`, if the equation is balancedWe can use `-1` as a constant to balance the function`=` `-int (-\text(sin) x)/(\text(cos) x)` Finally, integrate the function into a natural logarithm`-int (-\text(sin) x)/(\text(cos) x)` `=` `-ln (\text(cos) x)+c` `-ln (\text(cos) x)+c`
Quizzes
- Indefinite Integrals 1
- Indefinite Integrals 2
- Indefinite Integrals 3
- Indefinite Integrals of Exponential Functions
- Indefinite Integrals of Logarithmic Functions 1
- Indefinite Integrals of Logarithmic Functions 2
- Indefinite Integrals of Trig Functions
- Definite Integrals
- Definite Integrals of Exponential Functions
- Definite Integrals of Logarithmic Functions
- Definite Integrals of Trig Functions
- Areas Between Curves and the Axis 1
- Areas Between Curves and the Axis 2
- Area Between Curves
- Volumes of Revolution 1
- Volumes of Revolution 2
- Volumes of Revolution 3
- Trapezoidal Rule
- Simpsons Rule
- Applications of Integration for Trig Functions