Could you please help me integrate $$\int\frac{dx}{\sqrt{2+x-x^2}}$$ I am supposed to complete the square.. but I am seriously stuck. I have tried to square each side and stuff, but it does not seem to be working

Let $f\in L^1[a,b]$ satisfying $$\int^b_a t^kf(t) dt\,=0$$ for all positive integer $k$. Show that $f=0$ a.e. I did a similar problem where $\int^b_a t^kf(t) dt\,=0$ was true for all $k\in \Bbb{N} \cup \{0\}$. This is relatively easy and I did it using Weierstrass approximation theorem, but how to do when $k\in \Bbb{N}$ and k cannot […]

As my title says, I need help integrating $$\int _0^{\infty }\frac{e^{-t^2}-e^{-4t^2}}{t^2}dt$$ How would approach this? Thanks in advance.

I am trying to evaluate the following integral involving the Gauss Hypergeometric function, power, exponential and a Bessel Function: $$ \int_0^\infty x e^{-cx^2} {_2F_1(1,\frac{2} {ab},1+\frac{2} {ab},-zx^{-a})I_2(qx)dx} $$ where $a,b,c,z,q > 0 $ real numbers. Can you please provide some hints on solving this integral? Thank you for your time and patience

$I_n$ is given by $\int ^ {\pi/2}_{0} \sin^n(x) dx$ My attempt: I got to the fact that the statement is true iff $$\dfrac {n+1}{n+2} = \int ^ {\pi/2}_{0} \dfrac {\sin^2(x)}{\sin^n(x)}$$ I do not know how to prove the above statement using integration by parts.

This question already has an answer here: what is the$ \int \sin (x^2) \, dx$? 3 answers

let $\alpha'(x)=\beta(x), \beta'(x)=\alpha(x)$ and assume that $\alpha^2 – \beta^2 = 1$. how would I go about calculating the following anti derivative : $\int (\alpha (x))^5 (\beta(x))^4$d$x$. Thank you.

What is the integral of: $\int \frac{1}{5+3\sin x}dx$ My attempt: Using: $\tan \frac x 2=t$, $\sin x = \frac {2t}{1+t^2}$, $dx=\frac {2dt}{1+t^2}$ we have: $\int \frac{1}{5+3\sin x}dx= 2\int \frac 1 {5t^2+6t+5}dt $ I’ll expand the denominator: $5t^2+6t+5=5((t+\frac 3 5 )^2+1-\frac 1 4 \cdot (\frac 6 5)^2)=5((t+\frac 3 5)^2+0.64)$. So: $2\int \frac 1 {5t^2+6t+5}dt = \frac […]

I have to find out the complete integral of : $px^5-4q^3x^2+6x^2z-2=0$ My attempt:Let $f(x,y,z,p,q)=px^5-4q^3x^2+6x^2z-2$ So, $f_p=x^5,f_q=-12q^2x^2,f_x=5px^4-8q^3x+12xz,f_y=0,f_z=6x^2$ Applying Charpit’s method: $\frac{dx}{x^5}=\frac{dy}{-12q^2x^2}=\frac{dz}{px^5-12q^3x^2}=\frac{-dp}{5px^4-8q^3x+12xz+6px^2}=\frac{dq}{12x^2q^2} \implies \frac{dy}{-12q^2x^2}=\frac{dq}{12x^2q^2}\implies q=\sqrt{(y+c)}$ where $c$ is an arbitrary constant. I don’t see here a way to find out $p$. Please help me solve this problem .Even a hint would be much help. I’ll appreciate any […]

Find by integrating the area of the triangle vertices $$(5,1), (1,3)\;\text{and}\;(-1,-2)$$ I tried to make straight and integrate, but it is very complicated, there is some better way?

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