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Integral of a Trigonometric Function 2Integral of a Trigonometric Function 2
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Question 1 of 4
1. 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 2 of 4
2. 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` -
Question 3 of 4
3. Question
Find the integral$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\;\text{cos}x\;dx$$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)`First, integrate the trigonometric function$$\int_{\color{#00880A}{\frac{\pi}{4}}}^{\color{#9a00c7}{\frac{\pi}{2}}}\;\text{cos}\;x\;dx$$ `=` $$\bigg[\text{sin}\;x\bigg]_{\color{#00880A}{\frac{\pi}{4}}}^{\color{#9a00c7}{\frac{\pi}{2}}}$$ Integrate `\text(cos)x` Finally, get the difference of the upper and lower limits substituted to the integral as `x`.$$\bigg[\text{sin}\;x\bigg]_{\color{#00880A}{\frac{\pi}{4}}}^{\color{#9a00c7}{\frac{\pi}{2}}}$$ `=` $$\sin{\color{#9a00c7}{\frac{\pi}{2}}}-\sin{\color{#00880A}{\frac{\pi}{4}}}$$ Substitute the limits `=` `1-1/(sqrt2)` Evaluate `1-1/(sqrt2)` -
Question 4 of 4
4. Question
Find the integral$$\int_{0}^{\frac{\pi}{3}}\;3\;\text{sin}\frac{x}{2}\;dx$$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)`First, integrate the trigonometric function$$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}\;\text{sin}\;\frac{x}{2}\;dx$$ `=` $$\bigg[3\left(-\text{cos}\;\frac{x}{2}\right)\bigg]_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}$$ Integrate `3\text(sin) x/2` `=` $$\bigg[-6\text{cos}\;\frac{x}{2}\bigg]_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}$$ Simplify Finally, get the difference of the upper and lower limits substituted to the integral as `x`.$$\bigg[-6\text{cos}\;\frac{x}{2}\bigg]_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}$$ `=` $$-6\;\cos{\frac{\color{#9a00c7}{\frac{\pi}{3}}}{2}}-[-6\;\cos{\frac{\color{#00880A}{0}}{2}}]$$ Substitute the limits `=` `-6 \text(cos)(pi/6)+(6*1)` Evaluate `=` `-6*(sqrt3)/2+6` `\text(cos) pi/6=(sqrt3)/2` `=` `-3sqrt3+6` Simplify `-3sqrt3+6`
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