$\int\ e^{-x/2}\sqrt(sin-1)/(cosx+1)$  The result that I'm getting here contains a cosecx term but the answer has secx term! Please help.
121003
I need to learn how to find the definite integral of  the square root of a polynomial such as:      $$\sqrt{36x + 1}$$ or $$\sqrt{2x^2 + 3x + 7} $$  EDIT: It's not guaranteed to be of the same form. It could be any polynomial that can't be easily factored into squares.  This isn't homework, I'm studying for a final. And for context, I'm finding the arc length of a function.
68465
Integrate $\dfrac{dx}{(a^2 \cos^2x+b^2 \sin^2x)^2}$.  I can go up to the residue formula like in this example   but then I just can't end up with the result for when $n=2$. I keep messing up my math. Can someone please show me how to get to the answer when $n=2$? I think I've spent 3+ hours trying to get to it and no luck unfortunately. Thank you!
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?
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.
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 $dx=\dfrac{2\,dt}{1+t^2}$ $$\int \frac{1-t^2}{1+t-t^3}\,dt $$
37417
I want to solve the following integral without substitution:  $$\int{\sqrt{1-\sin2x}} \space dx$$    I have: $$\int{\sqrt{1-\sin2x}} \space dx = \int{\sqrt{1-2\sin x\cos x}} \space dx = \int{\sqrt{\sin^2x + \cos^2x -2\sin x\cos x}} \space dx$$    but this can be written in two ways: $\int{\sqrt{(\sin x - \cos x)^2}} \space dx$ or $\int{\sqrt{(\cos x - \sin x)^2} \space dx}$ and it seems to be pretty far from what the real result is:   Can I get any hints?  EDIT:  Thank you for your answers! So as you all showed, we have: $\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\sin x - \cos x)^2}} \space dx$ or $\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\cos x - \sin x)^2} \space dx} = \int{\sqrt{(-(\sin x - \cos x))^2} \space dx}$    combined: $\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\pm(\sin x - \cos x))^2} \space dx}$  so we actually have: $$\int{|\sin x - \cos x|} \space dx$$    I drew myself a trigonometric circle and if I concluded correctly, we have: 1. $\int{\sin x - \cos x} \space dx$ for $x \in [ \frac{\pi}{4} + 2k\pi , \frac{5\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$ 2. $\int{\cos x - \sin x} \space dx$ for $x \in [ -\frac{3\pi}{4} + 2k\pi , \frac{\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$    ... which means: 1. $x \in [ \frac{\pi}{4} + 2k\pi , \frac{5\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$ : $$\int{\sqrt{1-\sin2x}} \space dx = \int{\sin x - \cos x} \space dx = -\cos x - \sin x + C$$ 2. $x \in [ -\frac{3\pi}{4} + 2k\pi , \frac{\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$ : $$\int{\sqrt{1-\sin2x}} \space dx = \int{\cos x - \sin x} \space dx = \sin x + \cos x + C$$
43628
$$\int \sqrt{\sin x} ~dx.$$  Does there exist a simple antiderivative of $\sqrt{\sin x}$? How do I integrate it?
16406
I cant seem to work this out, can someone pls. help,  $$\int_{0}^{\frac{\pi }{2}}\frac{x dx}{\sin x +\cos x}$$  You are meant to use the property of integral that:   $$\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$$
94497
$$\int_{-\pi/4}^{\pi/4}\bigl(\cos x+ \sqrt {1+x^2}\sin^3x \cos^3x \bigr)\, dx $$  This question is from a math GRE practice test  I've tried to solve this integral for 2 days... starting to think it is a typo  The answer is $\sqrt2$   but I need to know how to solve with steps  Some useful trig identities and derivatives:  $$\frac{d}{dx} \sin^2x = 2\cos x \sin x$$  $$\frac{d}{dx} \tan^{-1}x = \frac{1}{1+x^2}$$  $$\sin^2x = \frac{1}{2} - \frac{1}{2} \cos 2x$$  $$\cos^2x = \frac{1}{2} + \frac{1}{2} \cos 2x$$  $$\sin 2x = 2\cos x \sin x$$
32634
Indefinite Integral with sin/cos  I can't find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$
60480
$$\displaystyle\int \:\sin^n\left(x\right)\cos^m\left(x\right)\mathrm  dx=\frac{\sin^{n+1}x\cos^{m-1}x}{m+n}+\frac{m-1}{m+n}\int \:\sin^nx\cos^{m-2}x\,\mathrm dx$$  I have been trying to solve for over a week now can someone please help me.
36902
How to find integral of: $$\int xe^x\sin(x)$$ I know it should be repeated integration by parts, but i do not know how many times i should do this and when to stop.
1954
$$\int \cos^4 x\,\mathrm dx  = \int (1 - \sin^2x)^2\,\mathrm dx.$$  I tried using $\cos^2(x) + \sin^2(x)$ = 1. This was on the integration by parts section of my textbook.  The integral I came out with is given me a hard time. I double checked my work and I feel I am on the right track but I feel stuck. I tried u substitution but there is no $\cos x$ to leverage with.
119613
Question:  $$\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}\mathrm dx.$$  What we did:  we tried using $t=\tan (\frac x2)$ and also dividing both numerator and denominator by $\sqrt {\cos x}$, eventually using the second method we got to this: $\displaystyle \int \frac {2t+2}{t^2+2t-1}-\frac {2}{t^2+2t-1} +\frac {\sqrt{2t(1-t^2)}}{t^2+2t-1}  $, for which we know how to solve the first and second integral but not the third...  Thanks
48006
Is it possible to get some help on this. I'm trying to help my friend solve this but it had been over 14 years since I did this type. All I have to work with is this.   Sorry if it is vague. Any help would be greatly appreciated!  Cosecant = cscx=1/sinx.
10597
The following integral can be obtained using the :   $$\int \frac{dx}{1+\cos^2 x} = \frac{\tan^{-1}(\frac{\tan x}{\sqrt{2}})}{\sqrt{2}}$$  Now assume we are performing this integration between $0$ and $2\pi$. Hence the result of the integration is zero.  On the other hand when looking at the integrand, $\displaystyle \frac{1}{1+\cos^2 x}$, we see that it is a periodic function that is never negative. The fact that it is never negative guarantees that the result of the integration will never be zero (i.e., intuitively there is a positive area under the curve).  What is going on here?
123795
Evaluate $\int \cos(\cos x)~dx$   I tried to use chain rule but failed. Can anyone help me please?