Other similar questions I have found used integration by part of alternative complicated transformations that I do not understand. Using the $\cos(x) = \frac{1}{2} * (e^{ix} + e^{-ix})$ identity, is there a way to find the solution, which should be $a/(a^2 + s^2)$?
123472
I need to find the Laplace transform of $\cos(at)$  I know that $L\{\cos(at)\}= \int_{0}^{\infty} e^{-st} \cos (at) dt$ but I am having trouble finding the integral  Thank you
93150
Evaluate the indefinite integral   $$\int\frac {\csc^2{x}-2005}{\cos^{2005}{x}} dx$$   I tried multiplying and dividing by $\sec^2 {x} $ and then setting $\tan{x}=y$ but no good. Then I set $\cos {x}=t $ and tried to create $\sin {x} $ in the numerator. But the integral which came was also a difficult one.    Please Help!     Thanks!     P.S. I am a high school student so kindly use elementary methods only.  Thanks again!
76497
$$\begin{align} \int \sec x \, dx  &amp;= \int \cos x \left(  \frac{1}{\cos^2x} \right)  \, dx \\ &amp;= \int \cos x \left(  \frac{1}{1-\sin^2x} \right) \, dx \\ &amp; = \int\cos x\cdot\frac{1}{1-\frac{1-\cos2x}{2}} \, dx \\ &amp;= \int \cos x \cdot\frac{2}{1+\cos2x} \, dx  \end{align}$$  I am stuck in here. Any help to integrate secant?
89755
How would I integrate the following?  $$\int_0^{\pi/2} x^2\sin(x)\,dx$$  I did $u=x^2$ and $dv=\sin(x)$  I got  $x^2-\cos(x)+2\int x\cos(x)\,dx.\quad$ I then used $u=x$ and $dv=\cos(x).$  I got   $$x^2-\cos(x)+2[x-\sin(x)-\int\sin(x)]$$  then  $x^2-\cos(x)+-2 \sin(x)(x)-\cos(x)\Big|_0^{\pi/2} =\dfrac{\pi^2}{4}-0-2$
123795
Evaluate $\int \cos(\cos x)~dx$   I tried to use chain rule but failed. Can anyone help me please?
96018
I need to solve this $$\int_{-\pi}^{\pi} \frac{e^{ixn}}{\sqrt{x^2+a^2}}\,dx,$$  where $i^2=-1$ and $a$ is a constant.
84451
I'm having trouble figuring out the Fourier series of $|\cos(x)|$ from $-\pi$ to $\pi$.  I understand its an even function, so all the $b_n$s are $0$  $$a_0 = \frac 2 \pi \int_0^\pi |\cos(x)|\,dx = 0$$  $$a_n = \frac 2 \pi \int _0^\pi  |\cos(x)| \cos(nx) \, dx = \frac 2 \pi \int_0^\pi \cos^2(x)\,dx.$$  since for all $j,k$ not equal the integral is zero.  so only $a_1$ remains. is this correct?   How would I evaluate $\sum_{n=1}^\infty (-1)^{n-1} /(4n^2 - 1)\  {}$?
111378
I think we use trig substitution for this. But I'm not sure where to even begin with this   $$\int \frac{25 - x^2}{x^4} \operatorname d\!x$$
97632
Find the value of $$\int\frac{1}{(x^2+x+1)^{1/2}} dx$$   Anyone can provide hint on how to integrate this, and how you know what method to use? (I mean, is there any general guideline to follow for solving?)  Thank you!
34118
$$\int_{0}^{\frac{\pi }{2}}x\cot(x)dx$$   I tried integration by parts and got $\dfrac{1}{2}\int_{0}^{\frac{\pi }{2}}x^{2} \csc^{2}x dx$ which doesn't help at all. I don't really know what to do. Any help will be greatly appreciated.
49455
Compute the definite integral      $$ \int^{\pi/2}_{0}\cos^{n}(x)\cos (nx)\,dx $$      where $n\in \mathbb{N}$.   My Attempt:  Using $\cos (x) = \frac{e^{ix}+e^{-ix}}{2}$, we get  $$ \begin{align} \int^{\pi/2}_{0}\cos^{n}(x)\cos (nx)\,dx&amp;=\int_{0}^{\pi/2} \left(\frac{e^{ix}+e^{-ix}}{2}\right)^n\left(\frac{e^{inx}+e^{-inx}}{2}\right)\,dx\\ &amp;= \frac{1}{2^n}\mathrm{Re}\left\{\int_{0}^{\pi/2} \left(e^{ix}+e^{-ix}\right)^n\cdot e^{inx}\,dx\right\}\\ &amp;=\frac{1}{2^n}\mathrm{Re}\left\{\int_{0}^{\pi/2}\left(e^{2ix}+1\right)^n\,dx\right\} \end{align} $$  Letting $z=e^{4ix}$ gives us  $$ \begin{align} e^{4ix}dx &amp;= \frac{1}{4i}dz\\ dx &amp;= \frac{dz}{4iz} \end{align} $$  So the integral becomes  $$\frac{1}{4\cdot 2^n}\mathrm{Re}\left\{\int_{C}\left(\sqrt{z}+1\right)\cdot \frac{dz}{iz}\right\}$$  How can I complete the solution from here?
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?
28096
How does one solve this equation?     $$\cos {x}+\sin {x}-1=0$$   I have no idea how to start it.  Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?  Thanks in advance!
30762
Question:     Find out $\displaystyle{\int e^{\cos x}~dx}$.   My Attempt:  Let $\cos x = y$. Hence $-\sin x\ dx = dy$ or $$dx = \displaystyle{\frac{-dy}{\sin x}=\frac{-dy}{\sqrt{1-\cos^2x}}=\frac{-dy}{\sqrt{1-y^2}}}$$ So  $$\begin{align}\int e^{\cos x}~dx &amp;= \int e^y\left(\frac{-dy}{\sqrt{1-y^2}}\right)\\ &amp;=-\int\frac{e^y}{\sqrt{1-y^2}}~dy \end{align}$$  This integral is one I can't solve. I have been trying to do it for the last two days, but can't get success. I can't do it by parts because the new integral thus formed will be even more difficult to solve. I can't find out any substitution that I can make in this integral to make it simpler. Please help me solve it. Is the problem with my first substitution $y=\cos x$ or is there any other way to solve the integral $\displaystyle{\int\frac{e^y}{\sqrt{1-y^2}}~dy}$?
6422
Integrate $$\int\frac{x^2dx}{1+x^4}$$ I've factored the denominator to $(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$ and got stuck.
81210
As the title states, I'm having quite a lot of trouble with the integral:  $$\int_{-\ln 2}^{\ln 2}  \frac{(\sin x+x)^{1/3}}{e^x}dx$$  The problem is that my standard (admittedly sparse) repertoire of tricks seems to have no effect at all. I can't think of any clever substitution that would work, and it looks like a nasty integral all-round.   I was hoping to apply some symmetry arguments (especially considering the limits of integration), but the exponential function is neither even nor odd, so I don't see how this would work.   I'm afraid I have absolutely no idea how to approach this.
75595
I have been attempting to solve this integral for several days in preparation for an exam and keep reaching dead ends.  I(a,b) = $\int_0^{\pi/2} \ln(a^{2}\cos^2(x)+b^{2}\sin^2(x)) dx$, where $a,b&gt;0$  My best approach so far is defining:  $F(x) = I(x, b)$  Then, I know:  $F(b) = I(b, b) = \pi/2 * \ln(b^2)$  Then I am now attempting to find $F'(x)$ using differentiation under the integral sign, and integrating from b to a.  This is where I am stuck - no matter what I do, I cannot solve the integral. Am I in the right direction or should this be solved in a different manner? I also considered something elliptical due to the acos, bsin in the original integral, yet it has also reached a dead end.