Integration problems involving simple substitutions

It might be helpful to see the basic integration formulas and how simple substitution is done, as discussed here, and here, since the following simple substitution problems are based on them.

Substitution is a very important concept in integration. If you want to learn integration, you need to learn substitution. Basically substitution is just spotting an appropriate expression in the given integral and then putting another variable in place of that expression. It involves a few rules, which will be clear from the following examples:

Example 1

`\int sin(x)cos(x) dx`
First, let us determine whether the above problem can be solved by using any of the basic integration formulas. There are integration formula for sin(x), and one for cos(x), but none for sin(x)cos(x). We are already familiar with basic substitution in integrals.

Now what we have to do is that we have to look for an expression in this integral, whose differential coefficient (or, simply, derivative) will give us the rest of the integral. Let us consider sin(x). The derivative of sin(x) is cos(x). Thus if we substitute
`t = sin(x)`
`dt = cos(x) dx`
So we can write the integral as,
`\int sin(x) cos(x) dx = \int t dt`
(if you are unable to understand this step then have a look at this post which discusses it in detail)
Now can you integrate `\int t dt` ? If you know the power rule, you can do this quite easily:
`\int t dt = t^2 / 2 + C`
The last step of any substitution, except possible in definite integrals, is to replace the original substitution. Otherwise, your answer is incorrect. Therefore,
`\int sin(x) cos(x) dx = sin^2(x) / 2 + C`     ... (Answer)

Example 2

 `\int sin(2x) dx`
This is quite easy. We know the formula `\int sin(x) dx = -cos(x) + C`, so we directly get `\int sin(2x) dx = -cos(2x) + C`, don't we? Actually no. The integration formula can not be directly applied if you have any expression in place of the single variable in the formula. Here we have `2x` in place of `x`. So the formula can't be used directly.

So we simply substitute
`t = 2x`.
Then dx is obtained by differentiating both sides of the equation with respect to x:
`dt = 2 dx`
`1/2 dt = dx`
So we replace 2x with 't' and `dx` with `1/2 dt` to get
`\int sin(t) 1/2 dt`
Numbers can be factored out of the integral (that is, if they are in multiplication),
`= 1/2 \int sin(t) dt`
Then applying the formula `\int sin(x) dx = -cos(x) + C`, we get
`= 1/2 (-cos(t)) + C`
Now replace t with 2x and simplfy to get:
`= -cos(2x)/2 + C`    ... (Answer)