Can you guys show me some different ways of using integration by parts to evaluate $\int \cos(\sqrt{x})\,dx$? I was trying with $u=\cos(\sqrt{x})$ and $dv=dx$, and it may have worked out eventually but things were getting messy. Maybe there's a substitution to be made before we assign $u,v$ to simple things. Thanks!
122468
My textbook says I should solve the following integral by first making a substitution, and then using integration by parts:  $$\int cos\sqrt x \ dx$$  The problem is, after staring at it for a while I'm still not sure what substitution I should make, and hence I'm stuck at the first step. I thought about doing something with the $\sqrt x$, but that doesn't seem to lead anywhere as far as I can tell. Same with the $cos$. Any hints?
72050
Evaluate the following integral:  $$\int \frac{\sqrt{\sin ^4x+\cos ^4x}}{\sin ^3x \cos x }dx$$ where $x \in \big(0,\frac{\pi}{2} \big)$  Could some give me hint as how to approach this question?  I tried to use the fact that $\sin ^4x+\cos ^4x=1-\frac{sin^22x}{2}$ but it didn't help. How should I proceed?
65890
How do you integrate $\int \frac{1}{a + \cos x} dx$?  Is it solvable by elementary methods?  I was trying to do it while incorrectly solving a homework problem but I couldn't find the answer.  Thanks!
57753
I was asked to solve this indefinite integral using Integration by parts.  $$\int \sqrt{1-x^2} dx$$  I know how to solve if use the substitution $x=\sin(t)$ but I'm looking for the Integration by parts way.   any help would be very appreciated.
32634
Indefinite Integral with sin/cos  I can't find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$
3572
The question:      Compute$$ \int \frac {x^2 \, \operatorname{d}\!x} {(x\sin x+\cos x)^2} $$   Tried integration by parts. That didn't work.  How do I proceed?
941
$$\int\sqrt{x^2+1}dx$$ I've been attempting this problem for days and I've made very little progress. I understand that I should substitute $x=\tan(x)$ and $dx=\sec^2(x)dx$ but beyond that I am very much lost.
96009
$$\int \frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}}$$  I tried substituting $x=z^2$ ... also $x=\tan^2 \theta$ ... but couldn't solve it either ways... if someone can help then it would be good.
42712
Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?   I have reduced this problem to $$   2\int_0^{\pi/2} \sqrt{\tan x} \ dx$$  but now, evaluating this integral is giving me some problems, simply substituting $u=\tan(x)$ and then $\mathrm{d}u=\sec^2(x)\mathrm{d}x \Rightarrow \frac{\mathrm{d}u}{1+u^2}=\mathrm{d}x$ and which in turn gives something a bit ugly, I was wondering which is the most elegant way to evaluate this?
23205
I am having real trouble evaluating this indefinite integral  $$\int{\frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}}}dx$$  Need I mention that I already tried WolframAlpha with little success? It returned a complicated expression full of many $\arctan$ terms.  Here is what I have already tried:   Seperating denominator as $(1-\sqrt{2}\sin{x}\cos{x})(1-\sqrt{2}\sin{x}\cos{x})$ and applying partial fractions. Multiplying and dividing by $\sec^4{x}$   Please provide any help. It would be appreciated if you give a hint rather than the whole solution.
43845
how i evaluate this integral with integration by substitution $x=\frac{\pi}{2}-t$  $$\int _0^{\frac{\pi }{2}}\:\frac{\sqrt[3]{\sin^2\left(x\right)}}{\sqrt[3]{\sin^2\left(x\right)}+\sqrt[3]{\cos^2\left(x\right)}}dx$$
40135
Make a substitution to express the integrand as a rational function and then evaluate the integral  $$\int {\frac{2}{x\sqrt{x+1}}}\, dx$$  What is the substitution that I have to make?
17869
$$\int \frac{\sin\left(x\right)}{1+x^2}dx$$  I have tried to integrate by parts but it doesn't work. How do I evaluate it? Any advice, hint or well-thought solution will be appreciated.
96462
I think I'm having a bad day. I was just trying to use integration by parts on the following integral:  $$\int \frac{dx}{x\log(x)}$$ Which yields $$\int \frac{dx}{x\log(x)} = 1 + \int \frac{dx}{x\log(x)}$$ Now, if I were to subtract  $$\int \frac{dx}{x\log(x)}$$ from both sides, it would seem that $0 = 1$. What is going on here?  Aside: I do know the integral evaluates to  $$\log(\log(x))$$ plus a constant.
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?
47738
While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you.      $$\int_0^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta.$$
3663
I've been thinking about integration lately, and I've come up with a question that I'm not sure how to address. Consider $$ \int \sin x\cos x \, \mathrm dx = - \int -\sin x \cos x \, \mathrm dx $$ I started with the integral on the left hand side, which suggests a typical $u$-substitution. Let $u=\sin x$ then $\, \mathrm du=\cos x \, \mathrm dx$. So the integral evaluates to $$ \int \sin x\cos x \, \mathrm dx = \frac{\sin^2(x)}{2} $$ But the original integral also suggests an alternate substitution. Let $u=\cos x$ and then $du=-\sin x \, dx$. So now $$ \int \sin x\cos x \, \mathrm dx =- \int -\sin x \cos x \, \mathrm dx= -\frac{\cos^2(x)}{2} $$ So now I have that the integral evaluates to two different functions. I've tried playing with some different trigonometric identities, but I haven't been able to show that this is true and I'm fairly certain I haven't had any success because the statement itself isn't true. What am I doing wrong? How do you evaluate $\int \sin x\cos x \, \mathrm  dx$?