U-substitution → Chain Rule
Last updated
Last updated
The
u-substitutionis to solve an integral of composite function, which is actually to UNDO theChain Rule.
Compare how we handle the composite functions with derivatives & integrals:
For taking the derivative of a COMPOSITE function, we apply the Chain rule.
For taking the integral of a COMPOSITE function, we apply the u-substitution.
We use u-substitution when we need to integrate an expression of the form of:
Strategy:
Find a function as u
Find or MAKE an u' at the outside so that you can pair u' with dx
Replace u' · dx with du, because u' = du/dx
Rewrite the Integral in term of u, and calculate the integral
Back substitute the function of u back to the result.
Selecting u is the most tricky part here.
Apparently, we ignore the wrapper sin() here.
We notice that the derivative of -x+2 is -1 which we could find it at outside.
So let u = -x+2 and u' = -1
So rewrite the integral to ʃ sin(u) · u' · dx = ʃ sin(u) · du
It looks quite neat, so the u = -x+2 is alright.
Apparently it's in form of ʃ u'/u · dx
So that we can make u'·dx = du and the integral becomes ʃ 1/u · du
Quite nice, so the answer would be out of there.
With a real quick eyeballing, we see it's in form of ʃ u' · u⁶ · dx
So with u' · dx = du we will get the simplified form ʃ u⁶ · du = u⁷/7
Back substitute function of u back to get the result:
Notice this radical form should directly use the Reversed Inverse Trig Rule:
So that we assume a = 1 & u = 3x.
Since u' = 3 so we need to make a 3 from nowhere.
Rewrite the formula to: 1/3 ʃ 3/(1+u²) ·dx = 1/3 ʃ 1/(1+u²) ·du
Apply the Reversed Inverse Trig Rule to get: 1/3 arctan(u) + C
Back substitute 3x to u and the boundaries back to x get the result π/6.
Solve:
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Solve:
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