My university mathematics is kind of poor. But as I am learning advanced mathematics this becomes a major shortcoming.  I tried to integrate $\int$$\sin^3\theta$$\cos^2\theta$d$\theta$ = $\int$(-3$\cos^3\theta$$\sin$$\theta$d$\theta$)$\left(-\frac13\right)$$\sin^2\theta$ = $\int$$\left(-\frac13\right)$($\sin^2\theta$)d($\cos^3\theta$) = $\left(-\frac13\right)$($\sin^2\theta$$\cos^3\theta$) - $\left(-\frac13\right)$$\int$$\cos^3\theta$d$\sin^2\theta$ = $\left(-\frac13\right)$($\sin^2\theta$$\cos^3\theta$) - $\int$(-2)$\cos^4\theta$d$\theta$ = $\left(-\frac13\right)$($\sin^2\theta$$\cos^3\theta$) + $\left(\frac2{15}\right)$$\cos^5\theta$  But this certainly is wrong. Why?
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
24025
I tried using integration by parts twice, the same way we do for  $\int \sin {(\sqrt{x})}$ but in the second integral, I'm not getting an expression that is equal to  $\int x\sin {(\sqrt{x})}$.  I let $\sqrt x = t$ thus, $$\int t^2 \cdot \sin({t})\cdot 2t dt =  2\int t^3\sin(t)dt = 2[(-\cos(t)\cdot t^3 + \int 3t^2\cos(t))] = 2[-\cos(t)\cdot t^3+(\sin(t)\cdot 3t^3 - \int 6t \cdot \sin(t))]]$$  which I can't find useful.
105361
This is about that old chestnut, $\newcommand{\d}{\mathrm{d}} \int \sin^2 x\,\d x$.   OK, I know that ordinarily you're supposed to use the identity $\sin^2 x = (1 - \cos 2x)/2$ and integrating that is easy. But just for the heck of it, I tried using the $u$-$v$ substitution method (otherwise known as integration by parts).  $$ \int \sin^2 x\,\d x = \int \sin x \sin x\,\d x $$  We can say $u=\sin x$ and $\d u=\cos x\,\d x$ while $\d v = \sin x\,\d x$ and $v = -\cos x$. When we put it all together:   $$\int \sin^2 x\,\d x = u v - \int v\,\d u = -\sin x \cos x - \int -\cos^2 x\,\d x$$  and doing the same routine with $\cos^2 x$ we get $u = \cos x$, $\d u = -\sin x\,\d x$, $\d v = \cos x\,\d x $ and $v = -\sin x$, leading to:  $$\begin{align}   \int \sin^2 x\,\d x = uv - \int v\,\d u     &amp;= -\sin x \cos x - (-\cos x \sin x - \int -\sin^2 x\,\d x) \\     &amp;= -\sin x \cos x + \sin x \cos x - \int \sin^2 x\,\d x \end{align}$$  which eventually works out to  $$2\int \sin^2 x\,\d x = 0$$  So I wanted to get an idea why this didn't work. Maybe it's higher math and the why will be beyond me (I would think that might be the case), or maybe it's one of those proofs that looks absurdly simple when shown that I am just unaware of.
98361
How can I calculate the following integral?  $$\int_0^t \cos(t-s)\sin(s)\ ds$$  I can't get the integral by any substitutions, maybe it is easy but I can't get it.
104093
$$\int{\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx}$$  $$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx=\int \frac{(1-\sin x)(2-\sin x)}{\sqrt{(1-\sin x)(2-\sin x)(1+\sin x)(2+\sin x)}}dx$$  I am stuck. Please help me....
3210
Cruising the old questions I came across juantheron asking for $\int \frac {\sec x\tan x}{3x+5}\,dx$ He tried using $(3x+5)^{-1}$ for $U$ and $\sec x \tan x$ for $dv$while integrating by parts.  below is his work.  How can I calculate $$ \int {\sec\left(x\right)\tan\left(x\right)  \over 3x + 5}\,{\rm d}x $$  My Try:: $\displaystyle \int \frac{1}{3x+5}\left(\sec x\tan x \right)\,\mathrm dx$  Now Using Integration by Parts::  We get  $$= \frac{1}{3x+5}\sec x +\int  \frac{3}{(3x+5)^2}\sec x\,\mathrm  dx$$  Here he hit his road block.   I tried the opposite tactic   Taking the other approach by parts.   let $$U= \sec x \tan x$$ then$$ du= \tan^2 x \sec x +\sec^3 x$$ and $$dv=(3x+5)^{-1}$$ then $$v=\frac 1 3 \ln(3x+5)$$ Thus $$\int \frac {\sec x \tan x}{3x+5}\,dx= \frac {\ln(3x+5)\sec x \tan x}{3} - \int \frac {\ln(3x+5) [\tan^2 x \sec x +\sec^3 x]}{3} \,dx$$    As you can see I got no further than he did.   So how many times do you have to complete integration by parts to get the integral of the original $\frac {\sec x \tan x}{3x+5} \, dx$ or is there a better way?
28612
I know $\int \tan^{1/3}\theta  d \theta$ is a non integrable but wolfram alpha does it easily. And can anyone explain how a fuction is non integrable? My argument is that there has to be a function which represents the area under a graph. It's just that we do not know it. Please shed a bit light on it. Also, how can we solve $\int \tan^{1/3}\theta  d \theta$?
1542
Calculating with Mathematica, one can have $$\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}\,\mathrm dt=\frac{\pi}{4}.$$   How can I get this formula by hand? Is there any simpler idea than using $u = \sin t$?     Is there a simple way to calculate $$ \int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}\,\mathrm dt $$ for $n&gt;3$? Could anyone come up with a reference for this exercise?
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?
71567
What is the result of this integration in a closed form?  $ \int{\cos{ x }^2}dx $  We can say $\int { cos\left( 2x \right) dx= } \frac { sin(x) }{ 2 } $  So why we can't say $\int { cos({ x }^{ 2 })dx=\frac { sin({ x }^{ 2 }) }{ 2x }  } $
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 $\tan(x)=t$.
69250
$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$  Put $x=\sin^2\theta,dx=2\sin \theta \cos \theta d\theta$  $\int_{0}^{\pi/2}\frac{\theta.2\sin \theta \cos \theta d\theta}{\sin^4\theta-\sin^2\theta+1}$ but this seems not integrable.Is this a wrong way?Is there a different way to solve it?
86644
We have to evaluate the following integral:  $$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$  I tried this:  I multiplied both the numerator and denominator by $\sec x$ And substituted $\tan x = t$.  But after that I got stuck.  The book where this is taken from gives the following as the answer: $$\ln(1+t)-\frac14\ln(1+t^4)+\frac1{2\sqrt2}\ln\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}-\frac12\tan^{-1}t^2+c$$ where $t=\sqrt{\cot x}$
24012
Here is a trigonometry problem.     Given   $$\frac{\cos(\alpha-3\theta)}{\cos^3(\theta)}=\frac{\sin(\alpha-3\theta)}{\sin^3(\theta)} = m$$   Show that    $$m^2+m\cos(\theta) = 2.$$   I tried to convert $\sin^3(x)$ into $\sin(3x)$ and similarly to cosine term, but couldn't get the answer. Please tell how to proceed further and any other way to solve it.
10042
Assume we have:  $$ \int (\cos{x} + \sin{x}\cos{x}) \, dx$$  Two ways to do it:  Use $$\sin{x}\cos{x} = \frac{ \sin{2x} }{2} $$ Then $$ \int \left(\cos{x} + \frac{\sin{2x}}{2} \right) \, dx = \sin{x} - \frac{ \cos{2x} }{ 4 } + C. $$ The other way, just see that $ u = \sin(x), du = \cos(x)dx $, $$ \int ( \cos{x} + \sin{x}\cos{x}) \, dx = \sin{x} + \frac{\sin^2{x}}{2} + C. $$  Now the part I don't see fully is, why aren't these results completely equal?  Taking the 2nd result, \begin{align} \sin{x} + \frac{\sin^2{x}}{2} &amp;= \sin{x} + \frac{1}{2} \, \left( \frac{1 - \cos{2x}}{2} \right) \\ &amp;= \sin{x} + \frac{1}{4} - \frac{\cos{2x}}{4}. \end{align}  So you have to absorb $\frac{1}{4}$ into $C$ for them to be equal.  Shouldn't they be equal straight away?
90593
Evaluate   $$ \int_{0}^{1} \arctan^{2}\left(\, x\,\right) \ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x $$     I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got    $$ -\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right) \over \cos^{2}\left(\,\theta\,\right)}\,{\rm d}\theta $$    After this, I thought of using the Taylor Expansion of $\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right)$ near zero but that didn't do any good.    Please Help!
105081
What I have done so far: Substituting $$x=3\sin(t)\Rightarrow dx=3\cos(t)dt$$ converting our integral to  $$I=\int\frac{x^3}{\sqrt{9-x^2}}dx=\int \frac{27\sin^3(t) dt}{3\sqrt{\cos^2(t)}}3\cos(t)dt\\ \Rightarrow \frac{I}{27}=\int \sin^3(t)dt=-\frac{\sin^2(t)\cos(t)}{3}+\frac{2}{3}\int\sin(t)dt\\ \Rightarrow I=18\cos(t)-9\sin^2(t)\cos(t)+c$$ Which is where I'm stuck. I can substitute back in for the $\sin(t)$ terms but don't know hot to deal with the $\cos(t)$ terms and back out information about $x$.   edited: because I miswrote the problem, I am sorry.