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In this video, we're going to go over a few indefinite integral problems. So, what is the integral of 4 dx? What is the answer for this problem? The anti-derivative of a constant, all you need to do is just add an x to it. This is going to be 4x. And you also need to add a a c value. Anytime you integrate a function, there's always going to be a constant that you need to add to it. Now the derivative of 4x is four. The derivative of any constant is zero. So that's why you always need to add the
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constant. So what about let's say the anti-derivative of pi? Let's say it's dy instead of dx. All you need to do is add a y variable to it. It's going to be pi * y + c. Now what about the anti-derivative of e dz? E is a constant, so it's just going to be E * Z plus C. Now, the next type of problem that you're going to see is when you need to integrate a variable raised to a constant. Let's say x raised to the n. This is equal to x raised to the n + 1 / n + 1 + c.
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So for example, let's say if we wish to find the anti-derivative of x^2 dx. This is equal to x 3r / 3 + c. The anti-derivative of x 3r is x 4 / 4 +
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c. Try this one. What is the anti-derivative of 8x cub? So, focus on x^ the 3 power. The 8 is just going to come along for the ride. The anti-derivative of x 3r is x 4 / 4. And now we can simplify. 8 / 4 is 2. So the final answer is 2x 4 +
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c. Try this one. What is the anti-derivative of 5x^ 6 power? So using the same technique, let's add one to the exponent and then divide by that result. So this is the answer. Now what about this expression? What's the anti-derivative of 7x dx? Well, there is an invisible one. So we can use the same technique. If we add one, it's going to be two and then divide by it plus c. So the anti-derivative of 3x is simply 3x^2 / 2 +
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c. What if we have a polomial function x^2 - 5x + 6? So you need to integrate each one separately. The anti-derivative of x^2 is x 3r / 3. And for 5x is going to be 5x^2 / 2. And then if you have a constant, just add a variable to it. So this is the answer for each of these problems. Before I begin, feel free to pause the video and work on it. So try this one. 4x + 8 x^2 - 9 dx. So the anti-derivative of x is x 4 / 4. And for x^2, it's x 3r / 3. And for the constant, simply add an x to it. And
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if you can reduce it, go ahead and do so. 4 / 4 is 1. So it's x 4 + 8/3 xqub - 9 x + c. So this is the answer. Now what about square root functions? What is the anti-derivative of the square root of x? If you see a question like this, rewrite it. This is the same as x to the 12. Now we need to add one to the exponent. 12 + 1 is the same as 12 + 2 over 2, which is 32. So this is going to be x to the 3s. And instead of dividing it by 3 halves, you can multiply by the reciprocal, which is
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2/3. So you can write the final answer as 2/3 x cub + c.
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Keep in mind there is an invisible two
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here. Here's another one that you can try. What is the anti-derivative of the cube root of x 4th? So go ahead and pause the video and work on this example. Now the first thing that we need to do is rewrite it. You can rewrite it as x^ 4/3. Now 4/3 + 1 is the same as 4 over 3 + 3 over 3, which is 7 over
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3. So once we add 1 to the exponent, it's just going to be x^ 7/3. Instead of dividing it by 7 over 3, let's multiply it by 3 over 7 and then add the plus c constant. So the final answer is 37 cube root x^ the 7th power plus
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c. Now what about this one? What's the anti-derivative of 3x - 1^ 2 dx? What would you do in this problem? The best thing to do is to foil this expression. 3x - 1^ 2 is 3x - 1 * 3x - 1 3x * 3x is 9 x^2 and then 3x * -1 that's -3x -1 * 3x is also -3x and finally we have -1 * 1 which is + 1. So now let's combine like terms. -3x + -3x is -6x. So this is what we now
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have. The anti-derivative of x^2 is x cub / 3. And for x to the 1st power is x^2 / 2. And for the constant add an x to it. So now let's simplify. 9 / 3 is 3. 6 / 2 is 3. And so this is the
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answer. Let's try this one. 2x + 1 * x - 2 dx. So just like the last example, we need to foil first. 2x * x is 2x^2. And then 2x * -2, that's -4x. 1 * x is x. And then 1 *
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-2. Now -4x + x is -3x. So now let's integrate it. This is going to be 2 x cub / 3 - 3x^2 / 2 - 2x + c. And this is the answer. We can't really simplify it. So we're going to leave it like
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that. Let's try this one. x 4 + 6 x cub / x dx. What should we do here? What would you do in this problem? Now, if you have a fraction with a single term in the bottom, separate it into two smaller fractions. So, x 4 / x is x cub. 6 x cub / x is 6 x^2. And now, as you can see, it's fairly easy to find the anti-derivative. So this is going to be x 4 / 4 + 6 xub / 3 + c which we can write it as 1/4 x 4 + 2 x cub +
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c. What about this one? What's the anti-derivative of 1x^2 dx? How can we integrate this function? For fractions like this, you want to rewrite it. If you move the x variable from the bottom to the top, the exponent is going to change sign. It's going to change from pos2 to -2. And now you can use the power rule. So if we add one to -2 and then divide by that result, this is going to be x^1 /1 + c. Now we can rewrite it. We can move the x back to the bottom. So the final answer is -1x +
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c. Now let's try another one like that. Try this one. 1x cub. So first let's rewrite it. This is x to the -3 dx. And now let's add one to the exponent. -3 + 1 is -2. Divide by that result. And you should get this. Now let's rewrite it. So we have a negative in front. We have a two in the bottom. Let's keep it there. And we're going to move the x to the bottom as well. So the -2 is going to change to pos2. So it's -1 / 2x^2 + c. Now let's try this one. 5 / x
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4th. So first let's rewrite it. This is 5x^ -4 and then let's add 1 to the exponent. So4 + 1 is -3 and then divide by3. And then we'll move the x back to the bottom. So it's -5 / 3 x cub +
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c. Now what is the anti-derivative of 1 /x? If we try to rewrite it and if we add one to the exponent, this is going to be 1 + 1 is zero. And if you have zero on the bottom, it's undefined. So this is not going to work for now. You just want to know that the anti-derivative of 1 /x is ln x. Likewise, let's say if you want to find the anti-derivative of 1 /x - 3. This is simply ln xus 3. The anti-derivative of 1 / let's say uh x + 4. This is just going to be ln x + 4.
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Now, what about this one? What's the anti-derivative of 5 x - 2? If you want, you can move the constant to the front. So, this expression is equivalent to 5 * 1x - 2 dx. And this portion is equal to ln x - 2 and just multiply by 5. So, this is the answer.
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Now let's move on to exponential functions. What is the anti-derivative of e to 4x? If it's e to the some number x to the 1st power. Here's what we need to do. It's just going to be e 4x / the derivative of 4x plus c. If it's like x^2 on top or x cub or something else, it won't work. It only works if the exponent is a linear function. So if you want to find the derivative of e^ 5x, it's simply going to be e 5x / 5. So what about e to x? It's going to be e x / the derivative
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of x, which is 1 plus c. So the anti-derivative of e to the x stays the same. It's just e to the x. Now let's try a few more examples. Let's try these two. 8 e to the 2x and also 12 e to the 3x. So this is going to be 8 e 2x / the derivative of 2x which is 2 and that reduces to 4 e 2x + c. For the last one, this is going to be 12 e^ 3x / 3 + c. And that reduces to 4 e^ 3x +
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c. Now let's move on to trig functions. What's the anti-derivative of cosine x dx? Now think backwards. The derivative of what function is cosine? The derivative of s is cosine. So the anti-derivative of cosine is positive sign. Now what is the anti-derivative of s? The derivative of cosine is negative sign. So the derivative of negative cosine is positive sign. Which means the anti-derivative of positive s is negative cosine. So let's say if we want to find the anti-derivative of cosine
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3x. This is going to be sin 3x. The angle has to stay the same but then divided by the derivative of 3x which is 3. This works is only if you have a linear function on the inside can you use this technique. So for example the anti-derivative of cossine 7x is simply sin 7x / 7 + c. Now what is the anti-derivative of 14 sin 2x? So this is just going to be 14. The anti-derivative of s is negative cosine 2x but divided by 2. So we can reduce that to -7 cossine 2x + c. 14 / 2 is 7.
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Let's try this one. 6 sin 3x dx. The anti-derivative of s is cosine 3x but / 3. So the final answer is -2 cosine 3x + c. 6 / 3 is 2. Now what is the anti-derivative of secant^ 2 dx? The derivative of tangent is squ. So the anti-derivative of squ is simply tangent x. So if we want to find the anti-derivative of 8^ 4x this is going to be 8 * tangent 4x but / 4 + c which becomes 2 tangent 4x +
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c. Now what is the anti-derivative of secant x tangent x? The derivative of secant is secant tangent. So this is equal to just sec x plus c. So if we have the anti-derivative of 12 secant 3x tangent 3x this is equal to 12 secant 3x / 3 plus c which is just uh 4 secant 3x plus c. Now what if you were to see an expression that looks like this? The what is the anti-derivative of x^2 sin x cub dx? What would you do in a problem like this? Now there's a technique called u substitution and you want to
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replace all the x variables with u variables. I'm going to make u equal to x cub. The reason being is the derivative of x cub is 3x^2 and the 3x^2 in this expression can cancel with the x^2 in that expression which is what we want. Now in the next step, solve for dx. du / 3x^2 is equal to dx. So what we're going to do is replace x cub with u and dx with du / 3x^2. So this is going to be x^2 sin u du over 3x^2. So notice that the x^2 cancels. And let's take this constant and move it to the front.
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So what we now have is 1/3 anti-derivative sin u du. And we know what the anti-derivative of s is. It's negative cosine. So this is 1/3 cosine u + c. Now at this point all you need to do is replace the u variable with what it was in the beginning x cub. So the final answer is 1/3 cossine x cub + c.
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Let's try another u substitution problem. Try this one. Do you think we should make u= x^2 + 3 or 5x? Notice that the derivative of x^2 will give you 2x, which can cancel the x and 5x. So you want to make u= x^2 + 3x. So du is going to be 2x dx. And then solve for dx. dx is du / 2x. Now we need to replace x^2 + 3 with u and dx with du over 2x. So this is going to be 5x raised to the u or time u raised to the 4th power and then time du / 2x. So we can cancel the x variable
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and the constant 5 / 2. Let's move it to the front. So let's put it on the left side of the integral. So this is 52 u 4 du. Now we can use the power rule. So if we add one to the exponent, it's going to be u 5 / 5 + c. So notice we can cancel these fives. So what we now have is 12 u to 5th + c which is let's make some space. At this point we can replace u with x^2 + 3. So the final answer is 12 x^2 + 3 raised to the 5th power + c. So this is it. Now let's try this problem. What is the
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anti-derivative of tangent x? You can either know this answer or you could find a way to get the answer. So what can we do now? tangent is sin / cosine. So in this form we could use u substitution. Let's replace u with cosine x. If we do that the derivative of cosine is going to be negative sign. And if we solve for dx, it's going to be du / s. Notice that the sign variable will cancel. So let's replace cosine with the u variable and let's replace dx with du /
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s. So the expression that we now have is the anti-derivative of -1 / u. If you recall the anti-derivative of 1 /x is ln x. So for 1 over u it's ln of u. Now we can replace the u variable with cosine. So what we now have is ln cossine x. Now there's a one in front of here. A property of natural logs allows you to take the coefficient and move it inside of ln. So this is going to be ln cossine x to the -1 power plus c which is the same as ln 1 / cosine x. Now 1 / cosine is secant. So the
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final answer is ln see x + c that is the anti-derivative of tangent
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x. Now what if you were to see this? What is the anti-derivative of x cosine x dx? We can't really use u substitution here. The derivative of cosine is negative sign. That's not going to cancel with the x. So, u substitution won't work. There's something else called integration by parts. And here's the formula. The integration of u dv is equal to uv minus the anti-derivative of v du. Let's make u equal to
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x. So this is the u part. dv we're going to make it equal to cosine x
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dx. Now we need to find du and v. du is the derivative of u. So the derivative of x is 1. v is the anti-derivative of dv. The anti-derivative of cosine is s. So this is going to be u * v. That's uh x * sin x minus the anti-derivative of v du which is simply sinx. Oh, let's not forget dx is here. du is 1 dx and udv is basically the original function. Now all we need to do is find the anti-derivative of s. The anti-derivative of s is negative cosine. So we have the final answer. It's going to be
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x sin x plus cosine x plus
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c. Let's try another integration by parts problem. Try this one. x e 4x dx. So you want to make u equal to x because when you find du, the x is going to disappear. D is going to be 1 DX. Now we're going to make DV equal to E 4X because we know how to find the anti-derivative of E 4X. E to X and sin X, they're basically repeating functions. The anti-derivative of E 4X is E 4X / 4. So using the formula, anti-derivative UV is UV minus anti-derivative V du. So uv is basically uh x * 1/4 e 4x.
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So that's 1/4 x e 4x minus the anti-derivative of v du which is simply 1/4 e 4x dx. So the anti-derivative of e 4x is just e 4x / 4 and then plus c. So now we can write the final answer which is 1/4 x e 4xus 4 * 4 is 16. So 1 16 e 4x + c. So that's integration by parts. Let's try this problem. What is the anti-derivative of 4 / 1 + x^2 dx? Now in this problem, we can't really use you substitution and we can't use integration by parts. So what can we do here? Now another technique that you can
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use is trigonometric substitution. It helps to know that 1 + tangent squ is equal to squ. So notice the expression 1 + x^2 that is an indication that we should make x= tangent theta. If x is tangent theta x^2 is going to be tangent and dx is the derivative of tangent which is dt. So let's replace x^2 with tang^ 2 and let's replace dx with squ theta d theta. So now 1 + tan^ 2 we know is 2 and the squares cancel in this problem. So what we now have is the anti-derivative of 4 d
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theta. The anti-derivative of 4 dx is just 4x. So 40 theta is simply 4 + c. So now we need to replace theta with something in terms of x. If x is equal to tangent theta, then the inverse tangent of x is equal to theta. Whenever you're dealing with an inverse function, you need to switch x and y. In this case, x and theta. So the final answer is 4 in tangent of x plus c. We just need to replace theta with inverse tan. So this is
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it. Let's try this one. Now sin^ 2 + cosine^ 2 is equal to 1. So 1 - sin^ 2 is equal to cosine^ square. So this is another trigonometric substitution. And based on that identity we want to make x= x= sin theta x^2 is sin^ 2 and dx is the derivative of s. It's going to be cosine theta d theta. So let's replace x^2 with sin^ 2. And let's replace dx with cosine theta d theta. So we know that 1 - sin^ 2 is equal to cosine^ 2 and the square root of cosine^ 2 is simply cosine
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theta. So at this point we can cancel the cosine function. So the expression that we have now is simply the anti-derivative of 3 d theta which is equal to 3 theta plus c. So now our last step is to replace theta with something. So if x is equal to sin theta what we need to do is take the inverse sign of both
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sides. The inverse sign of sin theta is simply theta. These two cancel. So inverse sin x is equal to
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theta. So the final answer is 3 inverse sin x + c.
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