# Integration by Substitution

Integration by Substitution:

Integration can be easily done by using  integration formulas , If the integral is in the standard form where we can easily apply formulas.

But if the function which is to be integrated is not in the standard form then it is either harder or impossible to use integration formulas to integrate.

In that case we need to to use Integration by Substitution method to integrate a given function.

In the method of Integration by substitution , We reduce a integral in non-standard form into a integral  in standard form by changing the variable into a new variable with suitable substitution.

Examples of Integration by substitution:
Example 1:
Let us try to integrate the integral: $\int x \times \sqrt{x-8} .dx$

The integral is not in the standard form and it is almost impossible to use integration formulas without doing suitable substitution.

If we analyze the integral then we can see that  we can substitute $\int x \times \sqrt{x-8} .dx$ by another variable lets say $y$.

So , $y = x - 8$

And , $x = y + 8$

Also , $\frac{dy}{dx} = 1$

So , $dy = dx$

Now we can rewrite the original Integral $\int x \times \sqrt{x-8}.dx$ as: $\int (y+8) \times \sqrt{y} . dy$

Now we can simplify the integral as: $\int y^{\frac{3}{2}} + 8 \times \int y^{\frac{1}{2}}$

Now we can easily use integration formulas and find it’s Antiderivate: $\int x \times \sqrt{x-8} .dx = \frac{2}{5} \times y^{\frac{5}{2}} +\frac{16}{3} \times y^{\frac{3}{2}}$

But we started from the variable “x” and got our answer in terms of variable “y” , which is not the desirable answer.

So we now again need to resubstitute the variable “y” with the variable “x” using our original substitution ( $y = x - 8$ )

And after using the integration by substitution method we finally get our answer: $\int x \times \sqrt{x-8} .dx = \frac{2}{5} \times ( x- 8)^{\frac{5}{2}} +\frac{16}{3} \times ( x- 8)^{\frac{3}{2}}$

Example 2:
Let us also try to ingrate the following integral by using integration by substitution method: $\int \frac{x}{4x^2 + 2} .dx$

Let us substitute $4x^2 + 2$ by $y$.

or , $y = 4x^2 + 2$

and , $\frac{dy}{dx} = 8x$

So , $dx = \frac{dy}{8x}$

And , $\int \frac{x}{4x^2 + 2} .dx = \int \frac{x}{y} . \frac{dy}{8x}$ $= \frac{1}{8} \ln y$ $= \frac{1}{8} \ln (4x^2 + 2)$

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