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### Simple substitution in integrals

All integrals can not be evaluated by directly using the integration formulae. Substitution is an important part of integration. It helps you to convert an integral into a 'nicer' form which will help you apply one or the other integration formulas, directly to it.

Substitution in integrals does not follow a rigid rule. Only by looking at the integral properly can you get to know what to substitute for what.

First, let us understand what substitution means. In integration, substitution means putting a variable in place of an expression. For example, in the following integral
\int (2x + 1)^10 dx
if we put t in place of 2x + 1 then it becomes quite simple to evaluate the integral because we get t^10, which can be evaluated using the power rule of integration.

But it is a rule in integration that the variable in 'dx' part of the integral should be the same as the variable of integration. Since we have changed the variable of integration from x to t by substituting t = 2x + 1, therefore we have to change dx to dt by some method.

The method to change dx to dt is to find the differential coefficient of the substitution. That is, we will differentiate both sides of the equation t = 2x + 1 to get:
d/dx t = d/dx (2x + 1)
dt/dx = 2
Then solve for dx,
dx =  1/2 dt
So now we know that we can write 1/2 dt in place of dx. The above integral, then, can be written as
\int t^10 1/2 dt
Factoring 1/2 out of the integral and applying the power rule of integration,
1/2 \int t^10 dt  ... (factoring out 1/2)
1/2 t^11 /11 + C   ... (applying power rule)
1/22 t^11 + C
Now comes the most important step: Replacing t with the original substitution 2x + 1.
1/22 (2x +1)^11 + C     ... (Answer)
Therefore we can write,
\int (2x + 1)^10 dx = 1/22 (2x +1)^11 + C
Remember that substitution in integrals does not follow a specific rule as to what you have to substitute for. Sometimes it may be apparent, and sometimes it may take time to spot the substitution, especially in bigger integrals. As you progress from simpler to more complex substitutions, gradually you will be able to do substitutions quite quickly.

For integrals in which you have to substitute linear expressions (all of the above examples are of that kind), you can remember a shortcut method to solve them quickly: The linear expression being substituted may have a coefficient with 'x'. For example coefficient of x is 2 in the expression 2x + 3. So what you have to do is just put t in place of the linear expression, write dt in place of dx, and integrate it. The important part is that you don't have to find 'dx', which saves you time. Then after finding the integral and replacing t with the original substitution, divide the whole integral by the coefficient of x you noted above.

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