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Question 1 of 4
Find the integral
∫ tan2x dx∫ tan2x dx
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Integrals of Trigonometric Functions
Integrating Trigonometric Functions
∫f(g(x))dx=f(g(x))⋅1g′(x)+c∫f(g(x))dx=f(g(x))⋅1g′(x)+c
First, convert the function into a derivable function
Take note that sec2=1+tan2xsec2=1+tan2x
sec2sec2 |
== |
1+tan2x1+tan2x |
sec2sec2 -1−1 |
== |
1+tan2x1+tan2x -1−1 |
Subtract 11 from both sides |
sec2-1sec2−1 |
== |
tan2xtan2x |
Therefore, we can use sec2-1sec2−1 as a derivable substitute
Finally, substitute the components into the formula
∫f(g(x))dx∫f(g(x))dx |
== |
f(g(x))⋅1g′(x)+cf(g(x))⋅1g′(x)+c |
|
∫sec2x−1dx∫sec2x−1dx |
== |
tanx−x+ctanx−x+c |
Substitute known values and integrate |
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Question 2 of 4
Find the integral
∫ tanx dx∫ tanx dx
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Integrals of Trigonometric Functions
Integrating Trigonometric Functions
∫f(g(x))dx=f(g(x))⋅1g′(x)+c∫f(g(x))dx=f(g(x))⋅1g′(x)+c
First, convert the function into a derivable function
Take note that tan x=sin xcos xtan x=sin xcos x
Therefore, we can use sin xcos xsin xcos x as a derivable substitute
Next, take note that ddxcos x=-sin xddxcos x=−sin x
This means that the function satisfies the derivative of a natural logarithm f′(x)f(x)f'(x)f(x), if the equation is balanced
We can use -1−1 as a constant to balance the function
|
== |
-∫ -sin xcos x−∫ −sin xcos x |
Finally, integrate the function into a natural logarithm
-∫ -sin xcos x−∫ −sin xcos x |
== |
-ln (cos x)+c−ln (cos x)+c |
-ln (cos x)+c−ln (cos x)+c
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Question 3 of 4
Find the integral
∫π2π4cosxdx∫π2π4cosxdx
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Integrals of Trigonometric Functions
First, integrate the trigonometric function
∫π2π4cosxdx∫π2π4cosxdx |
== |
[sinx]π2π4[sinx]π2π4 |
Integrate cosx |
Finally, get the difference of the upper and lower limits substituted to the integral as x.
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[sinx]π2π4 |
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= |
sinπ2−sinπ4 |
Substitute the limits |
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= |
1-1√2 |
Evaluate |
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Question 4 of 4
Find the integral
∫π303sinx2dx
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Integrals of Trigonometric Functions
First, integrate the trigonometric function
∫π30sinx2dx |
= |
[3(−cosx2)]π30 |
Integrate 3sin x2 |
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|
= |
[−6cosx2]π30 |
Simplify |
Finally, get the difference of the upper and lower limits substituted to the integral as x.
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[−6cosx2]π30 |
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= |
−6cosπ32−[−6cos02] |
Substitute the limits |
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= |
-6 cos(π6)+(6⋅1) |
Evaluate |
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= |
-6⋅√32+6 |
cos π6=√32 |
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= |
-3√3+6 |
Simplify |