## What is U substitution?

In mathematics, the U substitution is popular with the name integration by substitution and used frequently to find the integrals. So, you need to find an anti derivative in that case to apply the theorem of calculus successfully. This is the reason why integration by substitution is so common in mathematics. It could also be defined as the modified version of chain rule of differentiation where the function has been replaced by U and integrated later based on the fundamental integration or calculus formula.

#### U Substitution Formula

U substitution formula in mathematics could be given as below,

\[\large \int f\left(g\left(x\right)\right){g}’\left(x\right)dx=\int f\left(u\right)du\]

Where,u = g(x)

du = g′(x)dx

One more term to consider here is that u-substitution basically reverses the chain rule and it is simplifying the function of its anti derivative algebraically that could be recognized quickly. In simple words, u-substitution is a method for finding integrals. And the formula is needed to convert the one integral to another form that becomes easy to compute later. So, the formula will work left to right here or right to left in some cases to simplify the integrals.

The other name for this formula could be w substitution when used in a latter manner. This formula could also be used to find anti derivatives as well. When you need to find the relationship between x and u then you could find the relationship among derivatives as well by taking their differentiation and complete the substitution work later. An anti-derivative for the substituted function is possible to determine when the relationship among u and x variables is not maintained well.

**Question: **Evaluate the following integrals: \[\ \int \left(1-\frac{1}{w}\right)cos\left(w-1nw\right)dw\]

**Solution**

In this case we know how to integrate just a cosine so let’s make the substitution the stuff that is inside the cosine.

*u = w = 1nw*

\[\ du=\left(1-\frac{1}{w}\right)dw\]

So, as with the first example we worked the stuff in front of the cosine appears exactly in the differential. The integral is then,

\[\ =\int cos\left(u\right)du\]

\[\ =sin\left(u\right)+c\]

\[\ =sin\left(w-1nw\right)+c\]