So I was looking at how to obtain the following integral $$\int \sec(x)\,\mathrm dx$$ I saw that, to solve it, they divided and multiplied by $\tan(x)sec(x)/tan(x)sec(x)$. I understand that this is the equivalent to multiplying by $1$ and that they can do this. However, I would like to understand the logic behind why they are multiplying by this instead of some other trig ratio? Also is there any other way to solve this integral? I looked around but the only result I could find was multiplying and dividing by $\tan(x)sec(x)/tan(x)sec(x)$
33268
The standard approach for showing $\int \sec \theta \, \mathrm d \theta = \ln|\sec \theta + \tan \theta| + C$ is to multiply by $\dfrac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$ and then do a substitution with $u = \sec \theta + \tan \theta$. I like the fact that this trick leads to a fast and clean derivation, but I also find it unsatisfying: It's not very intuitive, nor does it seem to have applicability to any integration problem other than $\int \csc \theta \,\mathrm d \theta$.  Does anyone know of another way to evaluate $\int \sec \theta \, \mathrm d \theta$?
13485
I'm stuck with this integral  $\int\frac {\cos^4x}{\sin^3x} dx$    which I rewrote as  $\int \csc^3x \cos^4xdx$  then after using the half angle formula twice for $\cos^4x$ I got this  $\frac 14\int \csc^3x (1+\cos(2x))(1+\cos(2x))dx$   then after solving those products I got these integrals  $\frac 14 \{\int \csc^3xdx+2\int \csc^3x \cos(2x)dx + \int \csc^3x \cos^2(2x)dx\}$  I do know how to solve the $\int \csc^3xdx$ one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys!
114861
Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer.   $$\int (\cos^3x)(\sin^2x)dx  = \int(\cos x)(\cos^2x)(\sin^2x)dx  = \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$$  let $u = \sin x$, $du = \cos xdx$  $$\int(1-u^2)u^2du = \int(u^2-u^4)du  = \frac{u^3}{3} - \frac{u^5}{5} +C$$  Plugging in back $u$, we get $\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$ + C
3043
I want a hint on this following integral. I don't know the first step, do I do $u$-sub. or do I use trig. sub or something?  $$\int \frac{1}{2+ \sin x + \cos x} ~\mathrm{d}x$$
76330
I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ without using Weierstrass substitution, which is the usual technique.   When $a,b=1$ we can just multiply the numerator and denominator by $1-\cos x$ and that solves the problem nicely.   But I remember that the technique I saw was a nice way of evaluating these even when $a,b\neq 1$.
75214
How do I integrate the following $$\int \frac{dx}{\sin^3 x + \cos^3 x}$$ ?  It appears that I am supposed to break this up into $(\sin x + \cos x)(1-\cos x \sin x)$, but the next thing to do is not apparent to me.
19538
Evaluate $$\int \frac{1}{\sin x+\sec x}\,dx $$ Expressing $\sin x$ and $\cos x$ in terms of $\tan\frac{x}{2}$ i.e. putting $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$ and hence $dx=\dfrac{2\,dt}{1+t^2}$ $$\int \frac{1-t^2}{1+t-t^3}\,dt $$
67059
The question :  $$ \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx $$  I tried dividing by $\cos^2 x$ and splitting the fraction.  That turned out to be complicated(Atleast for me!)  How do I proceed now?
20215
Question says it all, how can I verify the following?  $$\int\sec x\ dx=\frac12 \ln \left|\frac{1+\sin x}{1-\sin x}\right| + C$$
65533
I was having a bit of trouble with an integral  $$\int {1 \over (x^2-1)^2}dx$$  It was given as a partial fractions problem, and I tried a substitution of  $x=sec(\theta);dx=sec(\theta)tan(\theta)d\theta$  to get $$\int{{\sec(\theta) \tan(\theta)}\over{\tan^4(\theta)}}d\theta=\int{\cos^2(\theta)\over \sin^3(\theta)}d\theta$$  but I can't seem to move past this. Any suggestions would be greatly appreciated.
9574
Integrate  $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$   I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$.
56865
Evaluate the following integral.$$\int \frac{\sec x}{\sqrt{3+\tan x}}dx$$  On putting $t=\tan x$ I am getting $$\int\frac{1}{\sqrt{(t^2+1)(t+3)}}dt$$.  How should i proceed from here.
66747
What is right way to calculate this integral and why?     $$ \int\frac{1}{2x}\text dx $$   I thought, that this substitution is right: $$ t = 2x $$ $$ \text dt = 2\text dx $$ $$ \frac{\text dt}{2} = \text dx $$ $$ \int\frac{1}{2x}\text dx=\int\frac{1}{t}\frac{\text dt}{2}=\frac{1}{2}\ln|2x| + C . $$  But it's not right, because this is the correct answer: $$ \int\frac{1}{2x}\text dx=\frac{1}{2}\int\frac{1}{x}\text dx=\frac{1}{2}\ln|x| + C . $$  Can someone explain me, why is the first way wrong? When I derive both results, I get the same result.
104646
How would I carry out the following integration? $$\int_0^12\arctan x^2$$  I tried to substitute $x^2 = \tan\theta$ From there I wrote the integral as: $$\int_0^{\pi/4}\theta\cdot \frac{\sec^2\theta}{\sqrt{\tan\theta}}\ d\theta$$  Now, is my only option to use integration by parts? Or is there a better method to solve this integral?
81688
We know that $$\int f(ax + b)\,\mathrm dx = \frac1aF(ax + b) + C$$  so while solving $$\int x^2\,\mathrm dx = \frac13x^3 + C$$  also, $$\int (x + 3)^4\,\mathrm dx = \frac15(x + 3)^5 + C$$  but for trig functions like,  $$\int \sin^2x\,\mathrm dx \neq \frac13\sin^3x + C$$  Instead, we have to simplify it first algebraically. So why does the intuition fail here?  Update: I seem to be confused with f. Till now $$\int \sin^2(1*x+0)\,\mathrm dx  \ $$, I have assumed f = (sin(x))^2 with a=1 and b=0. There seems something wrong with my understanding.
64923
$$\int \frac{dx}{\sqrt{x^2 -9}}$$  $x = 3 \sec \theta \implies dx = 3 \sec\theta \tan\theta d\theta$  $$\begin{align} \int \frac{dx}{\sqrt{x^2 -9}} &amp; = \frac{1}{3}\int \frac{3 \sec\theta \tan\theta d\theta}{\tan\theta} \\ \\ &amp; = \int  \sec\theta d\theta \\ \\ &amp; = \ln | \sec\theta + \tan\theta| + c\end{align}$$  $x = 3\sec \theta  \implies \sec\theta = \frac {x}{3}$  $\tan\theta = \frac{\sqrt{x^2 - 9}}{x}$  I have have confused x with 3 but I cannot get the proper answer which is  $$\ln | x +  \sqrt{x^2 - 9}| + c$$  I always get $\dfrac{x}{3}$ or $\dfrac{\sqrt{x^2 - 9}}{x}$ or some variation of that, I can't eliminate them to get their answer.
94284
This is a nice problem. I am trying to use nice methods to solve this integral, But I failed.  $$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx, $$  where  $\arctan{x}=\tan^{-1}{x}$  mark: this integral is my favorite one. Thanks to whoever has nice methods.  I have proved the following:  $$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\sum_{n=1}^{\infty}\dfrac{2^{n-1}H^2_{n-1}}{nC_{2n}^{n}}=\dfrac{\pi^3}{96}$$  where $$C_{m}^{n}=\dfrac{m}{(m-n)!n!},H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$  I also have got a few by-products  $$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=-I_{1}-2I_{2}$$  where $$I_{1}=\int_{0}^{1}\dfrac{\ln{(1-x^2)}}{1+x^2}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\dfrac{\pi}{4}\ln^2{2}+\dfrac{\pi^3}{32}-2K\times\ln{2}$$  and $$I_{2}=\int_{0}^{1}\dfrac{x\arctan{x}}{1+x^2}\ln{(1-x^2)}dx=-\dfrac{\pi^3}{48}-\dfrac{\pi}{8}\ln^2{2}+K\times\ln{2}$$  and same methods,I have follow integral $$\int_{0}^{1}\dfrac{\ln{(1-x^4)}\ln{x}}{1+x^2}dx=\dfrac{\pi^3}{16}-3K\times\ln{2}$$ where $ K $ denotes Catalan's Constant.