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Antiderivatives of Trig Functions 1Antiderivatives of Trig Functions 1
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Question 1 of 4
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 4
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 4
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 4
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`
Quizzes
- Antiderivatives (Indefinite Integrals) 1
- Antiderivatives (Indefinite Integrals) 2
- Antiderivatives (Indefinite Integrals) 3
- Antiderivatives of Exponential Functions
- Antiderivatives of Logarithmic Functions 1
- Antiderivatives of Logarithmic Functions 2
- Antiderivatives of Trig Functions 1
- Antiderivatives of Trig Functions 2
- Definite Integrals
- Definite Integrals of Exponential Functions
- Definite Integrals of Logarithmic Functions