Understanding \(u\) substitution

There are loads of techniques you can use to integrate functions, but this wiki is going to focus on one technique: \(u\) substitution.

THIS SECTION IS CURRENTLY ON PROGRESS

\(u\) substitution is a method where you can use a variable to simplify the function in the integral to become an easier function to integrate. This technique is actually the reverse of the chain rule for derivatives.

Evaluate the integral: \[\int { { (2-x) }^{ 10 } } dx\]

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Let us set \(u=2-x\).

Then \(\dfrac { du }{ dx } =-1\), \(du=-dx\) and \(dx=-du\). (It is extremely simple to find \(du\) if \(\dfrac { du }{ dx } \) is already known, so in future problems we will skip this step.)

Plugging \(u\) into our integral,

\[\int { { (2-x) }^{ 10 } } dx=\int { { u }^{ 10 }(-du) } =-\int { { u }^{ 10 }du } =-\frac { { u }^{ 11 } }{ 11 } +C=-\frac { { (2-x) }^{ 11 } }{ 11 } +C\]

Notice that the substitution made the integrand simpler, by making the problem into a problem where the power rule can be used.

Evaluate the integral: \[\int { { x }^{ 2 }{ ({ x }^{ 3 }+2) }^{ 101 } } dx\]

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Let us set \({ x }^{ 3 }+2\).

Then \(du=3{ x }^{ 2 }dx\) and \(dx=\dfrac { du }{ 3{ x }^{ 2 } } \).

Plugging \(u\) into our integral,

\[\int { { x }^{ 2 }{ ({ x }^{ 3 }+2) }^{ 101 } } dx=\int { { x }^{ 2 }{ u }^{ 101 } } \frac { du }{ 3{ x }^{ 2 } } =\frac { 1 }{ 3 } \int { { u }^{ 101 } } du=\frac { 1 }{ 3 } \frac { { u }^{ 102 } }{ 102 } +C=\frac { { ({ x }^{ 3 }+2) }^{ 102 } }{ 306 } +C\]

Notice that the substitution, again, made the integrand simpler, by making the problem into a problem where the power rule can be used.

Evaluate the integral: \[\int { \frac { 1 }{ \sqrt [ 4 ]{ 1-3x } } } dx\]

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Let us set \(u=1-3x\).

Then \(du=-3dx\) and \(dx=-\dfrac { du }{ 3 } \).

Plugging \(u\) into our integral,

\[\int { \frac { 1 }{ \sqrt [ 4 ]{ 1-3x } } } dx=\int { \frac { 1 }{ \sqrt [ 4 ]{ u } } } \left( -\frac { du }{ 3 } \right) =-\frac { 1 }{ 3 } \int { \frac { 1 }{ \sqrt [ 4 ]{ u } } } du=-\frac { 1 }{ 3 } \int { { u }^{ -\frac { 1 }{ 4 } } } du=-\frac { 1 }{ 3 } \frac { { u }^{ \frac { 3 }{ 4 } } }{ \frac { 3 }{ 4 } } +C=-\frac { 4\sqrt [ 4 ]{ { (1-3x) }^{ 3 } } }{ 9 } +C\]

Notice that the substitution, again, made the integrand simpler, by making the problem into a problem where the power rule can be used.

Evaluate the integral: \[\int { \sin ^{ 4 }{ (x) } \cos { (x) } } dx\]

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Let us set \(u=\sin { (x) } \).

Then \(du=\cos { (x) } dx\) and \(dx=\sec { (x) } du\).

Plugging \(u\) into our integral,

\[\int { \sin ^{ 4 }{ (x) } \cos { (x) } } dx=\int { { u }^{ 4 }\cos { (x) } \sec { (x) } } du=\int { { u }^{ 4 } } du=\frac { { u }^{ 5 } }{ 5 } +C=\frac { \sin ^{ 5 }{ (x) } }{ 5 } +C\]

Notice that the substitution made the integrand simpler, by making the problem into a problem where the derivative of \(\sin { (x) } \) can be used.

Here is an exercise for you: Evaluate the integral:

\[\int { \cos { (\pi x) } \cos { (\sin { (\pi x) } ) } } dx\]

Good Luck!