Linked Questions

So I came across with this integral today in my midterm: $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2 $$ And I got two correct answers: $$\frac {\sec^2(\pi x)}{4\pi} +C$$ And $$\frac {\tan^2(\pi x)}{...

While solving problems on indefinite integrals many a times I get answers which are different from those given in my text book's answer keys page. I then verify my solution steps to ensure that even ...

$$\int x(x^2+2)^4\,dx $$ When we do this integration with u substitution we get $$\frac{(x^2+2)^5}{10}$$ as $u=x^2+2$ $du=2x\,dx$ $$\therefore \int (u+2)^4\,du = \frac{(x^2+2)^5}{10} + C$$ ...

There seems to be no way to get these expressions from both the methods to match. I know the arbitrary constant could vary in different methods but even the strcture of both these expressions are not ...

For example, determine $\int \left(\frac{1}{2x+1}\right)dx$. Given that $f(x)$ = $\ln(2x+1)$ and $f'(x)$ = $\ln\left(\frac{2}{2x+1}\right)$. Would this be $\frac{1}{2} \int\left (\frac{2}{2x+1}\...

I'm going through Stewart's Calculus, and in section 7.3, example 7 asks for $\int \tan^3(x)dx$. In the book, the solution is $\frac{\tan^2(x)}{2} - \ln|\sec x| + C$. This was obtained in the ...

I tried Evaluating $\int \sqrt{\dfrac{5-x}{x-2}}dx$ using two different methods and got two different results. Getting two different answers when tried using two different methods:- M-$1$: $$\int \...

I want to find the integral $$\int\frac{x}{2x-2}dx$$ This is just a simple question from my textbook. But there seems to be two ways of solving it. If I simplify it to: $$\int1+\frac{2}{2x-2}dx$$ I ...

Evaluate the integral $$ \int\frac{a^x}{\sqrt{1-a^{2x}}}\ dx $$ In a recent question, the OP asked what the problem was with their method and why it did not match the reference book. I tried the ...

I am trying to integrate $\frac{1-x}{(x+1)^2}$, but I get to answers for two different methods: First, $\frac{1-x}{(x+1)^2} = \frac{1-x+1-1}{(x+1)^2} = \frac{2}{(x+1)^2} - \frac{x+1}{(x+1)^2} = \frac{...

I tried solving the following integral using integral by parts : $$\int \frac{x^3}{(x^2+1)^2}dx$$ but I got a different answer from Wolfram Calculator This is the answer that I got : $$\int \frac{x^3}{...

In looking at the equality $$\int \frac{a}{b(c-x)}dx = \int dt$$ I obtained different answers via different methods. Via one method, I got $$- \frac{a}{b} \ln(c-x) = t+C$$ Via another, I got $$- \frac ...

I was solving the integral $$\int\frac{\sin x}{\cos^3x}dx$$ I used the substitution $$u=\cos x \qquad du=-\sin x\,dx \qquad -du=\sin x dx$$ So the integral has the form $$\int-\frac{du}{u^3}=\frac{1}{...

I am currently taking a calculus class. My teacher while teaching a specific type of indefinite integral told one mysteriously beautiful of the solving the integral. The general form of the integral ...

If you compute the integral: $$ \int \frac{2x}{(x+1)^2}.$$ by substitution (using $u=x+1$, then you will get. $$ 2 \ln | 1 + x | + \frac{2}{1+x} + C.$$ But if, instead, you use integration by parts: $...

I have a doubt regarding basic integration $$ \int (1+x) \mathrm{d} x= \int 1 \mathrm{d}x+ \int x \mathrm{d} x =x+ \frac{x^2}{2} +c $$ But in another method it can also be done as $$ \int(...

When trying to integrate $\frac{1}{1+sin(x)}$, I used two methods. Weierstrass substitution gives me $\frac{-2}{\tan(\frac{x}{2}) + 1}$ Multiplying by the conjugate gives me $\tan x - \frac{1}{\cos ...

$$∫(2x+10)^{-1}dx$$ $$=\frac{1}{2}∫(x+5)^{-1}dx$$ $$=\frac{1}{2}\ln|x+5|+C$$ but here is the same integral, yet different answer $$∫(2x+10)^{-1}dx$$ Let $u = 2x+10\implies \frac{du}{dx} = 2 \implies ...

I have $\int \frac{e^{\ln(1+x)}}{1+x} dx$. Now, if I simplify $e^{\ln(1+x)}$ to $1+x$, I simply obtain $\int \frac{1+x}{1+x} dx= \int 1dx = x+C$. If I do not simplify and I apply the rule $\int f'(x)e^...

Find $\int-\csc^2x\cot xdx$ Using $\int f'f^{n}dx=\frac{f^{n+1}}{n+1}$ rule, I am getting two answers for the above question. Considering that the derivatives of $\csc x$ and $\cot x$ are $-\csc x\cot ...

$\displaystyle\int\frac{\mathrm dx}{10x-3}=$ ? Obviously, if we say $\,u=10x-3\,,\,$ then integration becomes $\displaystyle\frac{1}{10}\int\frac{\mathrm du}{u}=\frac {1}{10}\ln(10x-3)+c$ However, I ...

I just graduated high school and was doing some math in my spare time, but I can't seem to figure out why when I integrate this particular integral, $I = \int{\frac{x}{3-x}}dx$ I get different answers....

I was integrating the function $f(x)=\frac{x^2}{x-1}$ with two methods: 1) by starting with $u$-substitution and 2) by starting with polynomial long division. Upon my attempt at method 1, I let $u=x-1$...

After working $-2 \displaystyle\int (\cos^2\,x-\sin^2\,x)\,\sin\,2x\,dx$, I got $\frac{-\sin^2(2x)}{2}+c$. The answer I was given was $\frac{\cos^2(2x)}{2}+c$. Are these both legitimate solutions, ...

I integrated $e^x\sinh(x)$ by writing out $\sinh x$ as $\frac{e^x-e^{-x}}{2}$ and multiplying to get $\int$$\frac{e^{2x}-e^{0}}{2}dx$ which comes out to be $\frac{e^{2x}}{4}- \frac{x}{2}$ + C however, ...

I tried finding $\int{\dfrac{\cos{x}}{2+\sin{x}}}dx$ via Weierstraß substitution: $t = \tan{\frac{x}{2}}$ and alternatively via $u = 2 + \sin{x}$. The first approach results in a rational function in $...

I had my first integration test today. There, I met up with a integrand that gives two different functions as a whole. Taking $t=\tan x$:- $$\int \sec^2x\tan x dx = \int t \ dt=\frac{t^2}{2}+C=\frac{\...

I'm currently operating with the following integral: $$\int\frac{u'(t)}{(1-u(t))^2} dt$$ But I notice that $$\frac{d}{dt} \frac{u(t)}{1-u(t)} = \frac{u'(t)}{(1-u(t))^2}$$ and $$\frac{d}{dt} \frac{...

Given the very simple integral \begin{equation} \int -\frac{1}{2x} dx \end{equation} The obvious solution is \begin{equation} \int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx = -\frac{1}{2} ...

Answers obtained from two online integral calculators: $$\begin{align}\int\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}\,\mathrm dx &= -\sqrt{\dfrac{x + 1}{1 - x}} + \sqrt{\dfrac{x + 1}{1 - x}}x - 2\arcsin\...

In general, when doing integration by parts, shouldn't you get the same result no matter what "$u$" or "$\mathrm{d}v$" you pick? I understand the strategy for picking "$u$&...

$\newcommand{\d}{\mathrm{d}}$ Evaluate the integral using the indicated substituion. $$\int \cot x \csc^2x \,\d{x}, \qquad u= \cot x .$$ Differentiating both sides of $u$, then making the ...

I'm reading Thomas calculus on page 657. There is an integral of a function $$\frac{45}{2000-5t}$$ And I found the integration of $$\frac{45}{2000-5t}$$ is totally different from $$\frac{9}{...

I would like some help computing this integral: $$\int \sin(\pi t)\cos(\pi t)dt$$ Apparently, the answer should be $\frac{\sin^2(\pi t)}{2\pi}+C$ but I get $\frac{-\cos(2 \pi t)}{4 \pi}$+C instead. ...

If $f(x)=x-1$, I want to find $$ \int (x-1) \ dx $$. I make a variable substitution z = x-1, dx = dz so it becomes $$ \int z \ dz \\ = \frac{z^2}{2} + C $$. If I now substitute back $z=x-1,$ I get: $$ ...

Evaluate $$ \int\frac{\mathrm{d}x}{(1+x^2)\sqrt{1-x^2}}. $$ Set $x=\tan u\implies\mathrm{d}x=\sec^2u\,\mathrm{d}u$. $$ I=\int\frac{\mathrm{d}x}{(1+x^2)\sqrt{1-x^2}}=\int\frac{\sec^2u\,\mathrm{d}u}{\...

i know that $\int{\frac{1}{1+x²}dx}=\arctan(x)$ but a colleague showed me that we can do this trick : $$\int{\frac{1}{1+x²}dx}=\int{\frac{1}{(x+i)(x-i)}dx} =\frac{1}{2}i\int{\frac{1}{x+i}}dx-\frac{1}{...

So, after solving this question, I got two different answers? and I don't think it's supposed to be this way? Evaluate the integral $$\int \frac{x^2+2}{x+2} dx$$ Using polynomial long division, I ...

Evaluate: $$\int x(x^2-16)dx$$ I have noticed that this integral can be solved using two different methods, but I am not sure which one is the correct one. Way 1: Using $u$-subtitution Let $u=x^2-16, ...

This is my first post here and I searched the internet for things like "can u substitution yield an invalid result when finding an antiderivative?" and "manual integration and maxima produce different ...

I'm having trouble solving this problem. So I got that $\sqrt {1 + 2\sin(x)\cos(x)} = \sqrt{(\sin(x) + \cos(x))^2}$ using the Pythagorean Identity for 1, but then I get $\lvert \sin(x) + \cos(x) \...

I have calculated the indefinite integral $\int \frac{-\sin x}{\cos^3x}dx$ by two ways and I get different solutions. First, I substitute $u=\cos x$. Therefore, $$\int \frac{-\sin x}{\cos^3x}dx=\int u^...

So the Integral is $$\int \frac {dt}{t(t+1)^2}$$ So i thought of it two ways. 1.Substituting $\displaystyle y=\frac 1t$ Solution: $\displaystyle \ln\left|\frac{t}{1+t}\right|-\frac{t}{1+t}$ 2....

So the first is obvious: $$\int \frac{1}{2x+2} dx = \frac{1}{2}\ln(2x+2) + C$$ Now, what if I do something like this: $$\int \frac{1}{2x+2} dx = \int \frac{1}{2(x+1)} dx = (\frac{1}{2}\int \frac{1}{x+...

To take $$\int{\cot^5(x)}dx$$ I substitute $\csc(x)=u$, $du=-\csc(x)\cot(x)dx$ and then it seems like $$du = -u\cot(x)dx \quad\implies\quad \cot(x)dx=-\frac{du}{u} \tag1$$ is it not? but then $$\begin{...

I have to solve the following ODE $$y'=\frac{xe^{y}}{4x-1}$$ with $y(0)=0$. From the equation, I get $$\int e^{-y}dy=\int\frac{x}{4x-1}dx$$ In the second integral, if $u=4x-1\Rightarrow x=(u+1)/4$ and ...

Let a function be defined $f(x)=sin^4x+cos^4x,$ $x\in R$.Rewriting this, $$f(x)=(sin^2x+cos^2x)^2-2sin^2xcos^2x$$$$\implies f(x)=1-\frac{sin^22x}{2}$$$$\implies \frac{1}{2}\leq f(x)\leq1 $$ However,...

So here is my solving process: $$ \begin{split} f(x)=x\ln(x+\sqrt{1+x^2})-\sqrt{1+x^2}&\Rightarrow f'(x)=\ln(x+\sqrt{1+x^2})\\ &\Rightarrow f''(x)=\frac{1}{\sqrt{1+x^2}} \end{split} $$ and ...

One way to compute $\int \frac{dx}{a-x}$ for $a>0$ is $- \int \frac{d(a-x)}{a-x} = - \ln |a-x|$; another way is factoring out the $a$ in the denominator first: $$ \int \frac{\frac{1}{a}}{1 - \frac{...