# Solving an immediate indefinite integral of a composite function

• greg_rack
In summary: It's actually a little more complicated than that: both of those are only correct up to a constant. The correct way to do it is to say that$$\int \frac{1}{\sqrt{x^2 + a^2}} \ dx = \ln(x + \sqrt{x^2 + a^2}) + C$$which tells you that$$\int \frac{1}{\sqrt{x^2 + a^2}} \ dx = \ln(x + \sqrt{x^2 + a^2}) + C_1 = \ln (a\sqrt{\frac{x^2}{a^2} + 1}) + C_1 = \ln(ax + a\sqrt greg_rack Gold Member Homework Statement$$\int (\frac{1}{cos^2x\cdot tan^3x})dx$$Relevant Equations none That's my attempt:$$\int (\frac{1}{cos^2x\cdot tan^3x})dx = \int (\frac{1}{cos^2x}\cdot tan^{-3}x) dx$$Now, being ##\frac{1}{cos^2x}## the derivative of ##tanx##, the integral gets:$$-\frac{1}{2tan^2x}+c$$But there is something wrong... what? PeroK I'm missing what's supposed to be wrong here! greg_rack said: Homework Statement::$$\int (\frac{1}{cos^2x\cdot tan^3x})dx$$Relevant Equations:: none That's my attempt:$$\int (\frac{1}{cos^2x\cdot tan^3x})dx = \int (\frac{1}{cos^2x}\cdot tan^{-3}x) dx$$Now, being ##\frac{1}{cos^2x}## the derivative of ##tanx##, the integral gets:$$-\frac{1}{2tan^2x}+c$$But there is something wrong... what? Convert the denominator into a fraction with only sine and cosine factors. IOW, convert the ##\tan^3## factor. The resulting expression becomes a reasonably simple problem that can be done by nothing more complicated than substitution. PhDeezNutz Mark44 said: Convert the denominator into a fraction with only sine and cosine factors. IOW, convert the ##\tan^3## factor. The resulting expression becomes a reasonably simple problem that can be done by nothing more complicated than substitution. Is the OP's answer wrong? PeroK said: Is the OP's answer wrong? The correct answer is ##-\frac{1}{2sin^2x}+c## PeroK said: Is the OP's answer wrong? I didn't go through his work very closely -- I just showed an easier way to go about it. greg_rack said: The correct answer is ##-\frac{1}{2sin^2x}+c## That's what you got, isn't it? Mark44 said: Convert the denominator into a fraction with only sine and cosine factors. IOW, convert the ##\tan^3## factor. The resulting expression becomes a reasonably simple problem that can be done by nothing more complicated than substitution. I did it, and it took me to the correct answer. But then I found out the method explained in the OP which looks legit to me, but takes me to a wrong answer, and so I wanted to understand why PeroK said: That's what you got, isn't it? I got ##-\frac{1}{2tan^2x}+c## greg_rack said: I got ##-\frac{1}{2tan^2x}+c## Same thing. Don't forget the constant of integration! PeroK said: Same thing. Don't forget the constant of integration! What do you mean? How could ##tan^2x=sin^2x##? greg_rack said: What do you mean? How could ##tan^2x=sin^2x##?$$\frac{1}{tan^2x}= \frac{1}{sin^2x} + C$$For some constant ##C##. greg_rack and etotheipi @greg_rack it's because$$-\frac{1}{2\sin^2{x}} + c_1 = -\frac{1}{2} \csc^2 x + c_1 = -\frac{1}{2} (1+ \cot^2{x}) + c_1 = - \frac{1}{2\tan^2{x}} + c_2$$PhDeezNutz and greg_rack Thanks a lot guys! This is the same sort of situation you get with the integral ##\int \sin(x)\cos(x)dx##. You can do this in at least three different ways, of which I'll show two. 1. Let ##u = \sin(x)##, so ##du = \cos(x)dx##. The resulting antiderivative is ##-\cos^2(x) + C##. 2. Let ##u = \cos(x)##, so ##du = -\sin(x)dx##. The resulting antiderivative is ##\sin^2(x) + C## These results look different, but because ##\sin^2(x) + \cos^2(x) = 1##, ##\sin^2(x)## and ##-\cos^2(x)## differ only by a constant, 1. The third way involves writing the integrand as ##\sin(2x)##. greg_rack and PeroK greg_rack said: Thanks a lot guys! Everyone has to learn this lesson once! Usually it's:$$\int \sin (2x) dx = -\frac 1 2 \cos(2x) + C$$And$$\int \sin (2x) dx = \int 2 \sin x \cos x dx = -\cos^2 x + C$$Which are related by the trig identity$$\cos(2x) = 2\cos^2 x - 1$$greg_rack, PhDeezNutz and etotheipi another non-trig one that can sometimes catch people out$$\frac{1}{a} \ln{ax} + c_1 = \int \frac{\mathrm{d}x}{ax} = \frac{1}{a} \int \frac{\mathrm{d}x}{x} = \frac{1}{a} \ln{x} + c_2$$greg_rack and PeroK Another "favorite": In the integral ##\int\frac{1}{\sqrt{1-x^2}}dx##, depending on whether you substitute ##x=\sin(u)## or ##x=\cos(u),## you will get either ##\arcsin(x)+c_1## or ##-\arccos(x)+c_2.## greg_rack, PeroK and etotheipi$$\int \frac{1}{\sqrt{x^2 + a^2}} \ dx = \sinh^{-1}(\frac x a) + C_1 = \ln(x + \sqrt{x^2 + a^2}) + C_2

greg_rack and etotheipi

## 1. How do you solve an immediate indefinite integral of a composite function?

To solve an immediate indefinite integral of a composite function, you can use the chain rule and integration by substitution. First, identify the outer and inner functions of the composite function. Then, use the chain rule to find the derivative of the inner function. Finally, substitute the derivative into the integral and solve using standard integration techniques.

## 2. What is the chain rule and how does it apply to solving an immediate indefinite integral of a composite function?

The chain rule is a rule in calculus that allows you to find the derivative of a composite function. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function. In the context of solving an immediate indefinite integral of a composite function, the chain rule is used to find the derivative of the inner function, which is then substituted into the integral.

## 3. Can you provide an example of solving an immediate indefinite integral of a composite function using the chain rule?

Yes, for example, if we have the integral ∫(x+1)^2 dx, we can use the chain rule to solve it. First, we identify the outer function as (x+1)^2 and the inner function as x+1. The derivative of the inner function is 1, so we substitute it into the integral to get ∫(x+1)^2 dx = ∫1*(x+1)^2 dx. Then, we can solve the integral using standard integration techniques.

## 4. Are there any other techniques for solving an immediate indefinite integral of a composite function?

Yes, another technique is integration by substitution. This involves substituting a variable for the inner function and then using the chain rule to solve the integral. This technique can be particularly useful when the integral involves trigonometric or exponential functions.

## 5. Is it possible to solve an immediate indefinite integral of a composite function without using the chain rule or integration by substitution?

Yes, it is possible in some cases to solve the integral using other integration techniques such as integration by parts or partial fraction decomposition. However, the chain rule and integration by substitution are often the most efficient and straightforward methods for solving an immediate indefinite integral of a composite function.

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