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Popular Calculus >

integral from-pi to 0 of-picos(nx)

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Solution

∫−π0​−πcos(nx)dx

Solution

0
Solution steps
∫−π0​−πcos(nx)dx
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=−π⋅∫−π0​cos(nx)dx
Apply u-substitution
=−π⋅∫−πn0​ncos(u)​du
Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx=−πn1​⋅∫−πn0​cos(u)du
Use the common integral: ∫cos(u)du=sin(u)=−πn1​[sin(u)]−πn0​
Simplify −πn1​[sin(u)]−πn0​:−nπ​[sin(u)]−πn0​
=−nπ​[sin(u)]−πn0​
Compute the boundaries:0
=−nπ​⋅0
Simplify=0

Popular Examples

integral from 0 to pi of 1-cos(θ)integral from-1 to 1 of 9integral from 1 to 3 of (2x+1/x)integral from 0 to 8 of xsqrt(8-x)integral from 2 to 6 of (x^2-4)

Frequently Asked Questions (FAQ)

  • What is the integral from-pi to 0 of-picos(nx) ?

    The integral from-pi to 0 of-picos(nx) is 0
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