Calculus Examples

Evaluate the Integral integral from 0 to pi/2 of e^(sin(7pix))cos(7pix) with respect to x
0π2esin(7πx)cos(7πx)dx
Step 1
Let u2=sin(7πx). Then du2=7πcos(7πx)dx, so 17πdu2=cos(7πx)dx. Rewrite using u2 and du2.
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Step 1.1
Let u2=sin(7πx). Find du2dx.
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Step 1.1.1
Differentiate sin(7πx).
ddx[sin(7πx)]
Step 1.1.2
Differentiate using the chain rule, which states that ddx[f(g(x))] is f(g(x))g(x) where f(x)=sin(x) and g(x)=7πx.
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Step 1.1.2.1
To apply the Chain Rule, set u1 as 7πx.
ddu1[sin(u1)]ddx[7πx]
Step 1.1.2.2
The derivative of sin(u1) with respect to u1 is cos(u1).
cos(u1)ddx[7πx]
Step 1.1.2.3
Replace all occurrences of u1 with 7πx.
cos(7πx)ddx[7πx]
cos(7πx)ddx[7πx]
Step 1.1.3
Differentiate.
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Step 1.1.3.1
Since 7π is constant with respect to x, the derivative of 7πx with respect to x is 7πddx[x].
cos(7πx)(7πddx[x])
Step 1.1.3.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
cos(7πx)(7π1)
Step 1.1.3.3
Simplify the expression.
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Step 1.1.3.3.1
Multiply 7 by 1.
cos(7πx)(7π)
Step 1.1.3.3.2
Move 7 to the left of cos(7πx).
7cos(7πx)π
Step 1.1.3.3.3
Reorder the factors of 7cos(7πx)π.
7πcos(7πx)
7πcos(7πx)
7πcos(7πx)
7πcos(7πx)
Step 1.2
Substitute the lower limit in for x in u2=sin(7πx).
ulower=sin(7π0)
Step 1.3
Simplify.
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Step 1.3.1
Multiply 7π0.
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Step 1.3.1.1
Multiply 0 by 7.
ulower=sin(0π)
Step 1.3.1.2
Multiply 0 by π.
ulower=sin(0)
ulower=sin(0)
Step 1.3.2
The exact value of sin(0) is 0.
ulower=0
ulower=0
Step 1.4
Substitute the upper limit in for x in u2=sin(7πx).
uupper=sin(7ππ2)
Step 1.5
Simplify.
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Step 1.5.1
Multiply 7ππ2.
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Step 1.5.1.1
Combine π2 and 7.
uupper=sin(π72π)
Step 1.5.1.2
Combine π72 and π.
uupper=sin(π7π2)
Step 1.5.1.3
Raise π to the power of 1.
uupper=sin(7(π1π)2)
Step 1.5.1.4
Raise π to the power of 1.
uupper=sin(7(π1π1)2)
Step 1.5.1.5
Use the power rule aman=am+n to combine exponents.
uupper=sin(7π1+12)
Step 1.5.1.6
Add 1 and 1.
uupper=sin(7π22)
uupper=sin(7π22)
Step 1.5.2
Evaluate sin(7π22).
uupper=0.01390333
uupper=0.01390333
Step 1.6
The values found for ulower and uupper will be used to evaluate the definite integral.
ulower=0
uupper=0.01390333
Step 1.7
Rewrite the problem using u2, du2, and the new limits of integration.
00.01390333eu217πdu2
00.01390333eu217πdu2
Step 2
Combine eu2 and 17π.
00.01390333eu27πdu2
Step 3
Since 17π is constant with respect to u2, move 17π out of the integral.
17π00.01390333eu2du2
Step 4
The integral of eu2 with respect to u2 is eu2.
17πeu2]00.01390333
Step 5
Substitute and simplify.
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Step 5.1
Evaluate eu2 at 0.01390333 and at 0.
17π((e0.01390333)-e0)
Step 5.2
Simplify.
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Step 5.2.1
Anything raised to 0 is 1.
17π(e0.01390333-11)
Step 5.2.2
Multiply -1 by 1.
17π(e0.01390333-1)
17π(e0.01390333-1)
17π(e0.01390333-1)
Step 6
The result can be shown in multiple forms.
Exact Form:
17π(e0.01390333-1)
Decimal Form:
0.00063663
0π2esin(7πx)cos(7πx)dx
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