Calculus Examples

Evaluate the Integral integral from 0 to pi/3 of square root of 1+tan(x)^2 with respect to x
π301+tan2(x)dx
Step 1
Simplify.
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Step 1.1
Rearrange terms.
π30tan2(x)+1dx
Step 1.2
Apply pythagorean identity.
π30sec2(x)dx
Step 1.3
Pull terms out from under the radical, assuming positive real numbers.
π30sec(x)dx
π30sec(x)dx
Step 2
The integral of sec(x) with respect to x is ln(|sec(x)+tan(x)|).
ln(|sec(x)+tan(x)|)]π30
Step 3
Simplify the answer.
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Step 3.1
Evaluate ln(|sec(x)+tan(x)|) at π3 and at 0.
ln(|sec(π3)+tan(π3)|)-ln(|sec(0)+tan(0)|)
Step 3.2
Simplify.
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Step 3.2.1
The exact value of sec(π3) is 2.
ln(|2+tan(π3)|)-ln(|sec(0)+tan(0)|)
Step 3.2.2
The exact value of tan(π3) is 3.
ln(|2+3|)-ln(|sec(0)+tan(0)|)
Step 3.2.3
The exact value of sec(0) is 1.
ln(|2+3|)-ln(|1+tan(0)|)
Step 3.2.4
The exact value of tan(0) is 0.
ln(|2+3|)-ln(|1+0|)
Step 3.2.5
Add 1 and 0.
ln(|2+3|)-ln(|1|)
Step 3.2.6
Use the quotient property of logarithms, logb(x)-logb(y)=logb(xy).
ln(|2+3||1|)
ln(|2+3||1|)
Step 3.3
Simplify.
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Step 3.3.1
2+3 is approximately 3.7320508 which is positive so remove the absolute value
ln(2+3|1|)
Step 3.3.2
The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.
ln(2+31)
Step 3.3.3
Divide 2+3 by 1.
ln(2+3)
ln(2+3)
ln(2+3)
Step 4
The result can be shown in multiple forms.
Exact Form:
ln(2+3)
Decimal Form:
1.31695789
 [x2  12  π  xdx ]