Hi everyone, we are now in module 5. It's all about anti-differentiation by substitutions. At the end of the learning episode, you are expected to compute the antiderivative of a function using the substitution rule. Now, let us have a recap. Let us recall first the following antiderivatives of trigonometric functions. We have the antiderivative of sine x dx is equal to negative cosine x plus c.
Then we have the antiderivative of cosine x dx is equal to sine x plus c. The next one is the antiderivative of second squared x dx is equal to tangent x plus c. Then we have the antiderivative of cosecond squared x dx is equal to negative cotangent x plus c. And for antiderivative of second x tangent x dx is equal to second x plus c. And the last one, the antiderivative of cosecant x cotangent x dx is equal to negative cosecant x plus c.
Integration by substitution, also called u-substitution, is a method to find an integral. but only when it can be set up in a special way. The first and most vital step is to be able to write our integral in this form, the antiderivative of f of g of x times g prime of x dx.
Note that we have g of x and its derivative g prime of x dx. Like in the given example number 1, the antiderivative of quantity x minus 1 raised to 3 dx. Okay, let us see.
The derivative of x minus 1 is 1. So we have the antiderivative of the quantity of x minus 1 raised to 3 dx is equal to the antiderivative of the quantity of x minus 3 raised to 3 times 1 dx. In this case, the antiderivative of the quantity x minus 1 raised to 3 times 1 dx. So x minus 1 is our u here raised to 3 and 1 dx is our du. So we can now integrate the antiderivative of quantity u raised to 3 du is equal to the quantity of u raised to 3 plus 1 all over 3 plus 1 plus c equals u raised to 4 all over 4 plus c now put u which is equal to x minus 1 back again therefore the antiderivative of quantity x minus 1 raised to 3 dx is equal to quantity x minus 1 raised to 4 all over 4 plus c Let us have another example, number 2. The antiderivative of the quantity of x raised to 5 minus 3 raised to 8 times 5x raised to 4 dx.
Let u is equal to x raised to 5 minus 3. So, our u is x raised to 5 minus 3. And the derivative of u, du is equal to du 5x to the 4th dx. Okay. So, this is the given.
The antiderivative of the quantity of x raised to 5 minus 3 raised to 8. Times 5x raised to 4 dx. We have now the antiderivative of the quantity of u raised to 8 du. So our u here is the quantity of x raised to 5 minus 3 and our du is 5x to the 4th dx.
Let us now integrate. The antiderivative of the quantity of u raised to 8 du. is equal to the quantity of u raised to 8 plus 1 all over 8 plus 1 plus c is equal to u raised to 9 all over 9 plus c. Now put u which is equivalent to quantity x raised to 5 minus 3 and du which is equal to 5x raised to 4 dx back again. Therefore the antiderivative.
of the quantity of x raised to 5 minus 3 raised to 8 times 5x to the 4th dx is equal to the quantity of x raised to 5 minus 3 raised to 9 all over 9 plus c all right so the next example is the antiderivative of the quantity of 5x minus 2 raised to 7 dx. Here let u is equal to 5x minus 2 and the derivative of u or du is equal to 5 dx. Now dividing both sides by 5 we have now one-fifth du is equal to dx. All right so once again the given is the antiderivative. of the quantity of 5x minus 2 raised to 7 dx our u here is 5x minus 7 and our dx is 1 5th du so we have now 1 5th times the anti-derivative of u raised to 7 du we can now integrate 1 5th Times the antiderivative of u raised to 7 du will give us 1 fifth times u raised to 7 plus 1 all over 7 plus 1 plus c.
Equals 1 fifth times u raised to 8 all over 8 plus c. Now put u which is equivalent to quantity 5x minus 2 back again and simplify. So we have now the antiderivative of the quantity of 5x minus 2 raised to 7 dx is equal to 1 fifth times the quantity of 5x minus 2 raised to 8 all over 8 plus c.
Therefore, the antiderivative of the quantity of 5x minus 2 raised to 7 dx is equal to the quantity of 5x minus 2 raised to 8. all over for t plus c. Alright, the next example. The antiderivative of 2x all over the square root of x squared plus 1 dx. Here, let u is equal to x squared plus 1 and our du, the derivative of u is 2x. dx.
So the antiderivative of 2x all over the square root of x squared plus 1 dx can also be written in the form of the antiderivative of 1 all over the square root of x squared plus 1 times 2x dx. So we have now the antiderivative of 1 all over the square root of u times du. can be written in the form of the antiderivative of u raised to negative one half du to be able to integrate we have the antiderivative of u raised to negative one half du equals quantity u raised to negative one half plus one all over negative one half plus one plus c so we have now quantity u raised to one half all over one half plus c equals quantity u raised to one half times the reciprocal of one half which is equal to two over one plus c equals two times the quantity of u raised to one half plus c now put u which is equal to quantity x squared plus one and du which is equal to 2x dx. Back again and simplify.
Therefore, we have now the antiderivative of 2x all over the square root of x squared plus 1 dx. It's equal to 2 times the square root of x squared plus 1 plus c. Alright?
There are activities provided in the module for you to check your understanding about this lesson. You may try answering them after this learning episode. According to Gordon B. Hinckley, there is no substitute under the heavens for productive labor. It is the process by which dreams become realities. It is the process by which idle visions become dynamic achievement.
God bless and keep safe always.