# Solving an immediate indefinite integral of a composite function

• greg_rack
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$$\int \frac{1}{\sqrt{x^2 + a^2}} \ dx = \ln(x + \sqrt{x^2 + a^2}) + C_1 = \ln (a\sqrt{\frac{x^2}{a^2} + 1}) + C_1 = \ln(ax + a\sqrtf #### greg_rack Gold Member Homework Statement$$\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$$But there is something wrong... what? PeroK I'm missing what's supposed to be wrong here! Homework Statement::$$\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$$But there is something wrong... what? Convert the denominator into a fraction with only sine and cosine factors. IOW, convert the ##\tan^3## factor. The resulting expression becomes a reasonably simple problem that can be done by nothing more complicated than substitution. PhDeezNutz Convert the denominator into a fraction with only sine and cosine factors. IOW, convert the ##\tan^3## factor. The resulting expression becomes a reasonably simple problem that can be done by nothing more complicated than substitution. Is the OP's answer wrong? Is the OP's answer wrong? The correct answer is ##-\frac{1}{2sin^2x}+c## Is the OP's answer wrong? I didn't go through his work very closely -- I just showed an easier way to go about it. The correct answer is ##-\frac{1}{2sin^2x}+c## That's what you got, isn't it? Convert the denominator into a fraction with only sine and cosine factors. IOW, convert the ##\tan^3## factor. The resulting expression becomes a reasonably simple problem that can be done by nothing more complicated than substitution. I did it, and it took me to the correct answer. But then I found out the method explained in the OP which looks legit to me, but takes me to a wrong answer, and so I wanted to understand why That's what you got, isn't it? I got ##-\frac{1}{2tan^2x}+c## I got ##-\frac{1}{2tan^2x}+c## Same thing. Don't forget the constant of integration! Same thing. Don't forget the constant of integration! What do you mean? How could ##tan^2x=sin^2x##? What do you mean? How could ##tan^2x=sin^2x##?$$\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$$PhDeezNutz and greg_rack Thanks a lot guys! This is the same sort of situation you get with the integral ##\int \sin(x)\cos(x)dx##. You can do this in at least three different ways, of which I'll show two. 1. Let ##u = \sin(x)##, so ##du = \cos(x)dx##. The resulting antiderivative is ##-\cos^2(x) + C##. 2. Let ##u = \cos(x)##, so ##du = -\sin(x)dx##. The resulting antiderivative is ##\sin^2(x) + C## These results look different, but because ##\sin^2(x) + \cos^2(x) = 1##, ##\sin^2(x)## and ##-\cos^2(x)## differ only by a constant, 1. The third way involves writing the integrand as ##\sin(2x)##. greg_rack and PeroK Thanks a lot guys! Everyone has to learn this lesson once! Usually it's:$$\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