Q: How to integrate this type of fractions $$\int^{\frac{\pi}{2}}_{\frac{\pi}{3}} \frac{\sin\frac{\theta}{2}}{1+\sin\frac{\theta}{2}} d\theta $$ What should I do here? I don’t even know where to start from. Please help me by giving me a hint.

This problem (if my derivations of them are correct) lead me to calculate the following integrals: $$I_1 = \int_0^2 { \int_0^{\frac{1}{2}x_1} {\frac{-1+x_1x_2-2x_2}{x_1-2x_2}} }dx_2dx_1$$ $$I_2 = \int_0^2 { \int_{\frac{1}{2}x_1}^1 {\frac{1-x_1x_2+2x_1-2x_2}{x_1-2x_2}} }dx_2dx_1$$ $$I_3 = \int_0^2 { \int_0^{\frac{1}{2}x_1} {\frac{-1-x_1x_2+2x_1-2x_2}{x_1-2x_2}} }dx_2dx_1$$ $$I_4 = \int_0^2 { \int_{\frac{1}{2}x_1}^1 {\frac{1+x_1x_2-2x_2}{x_1-2x_2}} }dx_2dx_1$$ How can this be done?

I need to prove this following double integral $$\int_{0}^{\infty} \left(\int_{0}^{t}f(s)\:g(t-s)\:ds \right)dt$$ can be rewritten like this: $$\int_{0}^{\infty}f(s)\:ds\:\int_{s}^{\infty}g(t-s)\:ds$$ I tried a variable change but I didn’t go too far. $f(x)$ and $g(x)$ are continuous functions. Thank you in advance.

If $f\in R[a,b]$, then prove that $\lim_{h\to0}\int_{c}^{d}|f(x+h)-f(x)|dx=0$, where $a<c<d<b$. $R[a,b]$ is the set of all Riemann Integrable functions on $[a,b]$. I think the well-known step function corresponding to the lower Riemann sum may work but I am having some doubts. Indeed, given any $\varepsilon>0$ because of integrability of $f$ I can find a partition $P$ […]

Question: Let $$F(x) =\int_{x^3}^{5}(cos^2t-te^t)dt $$ Find $F'(x)$ We were not explicitly taught about this during the semester but from what I can gather from online readings is that $$F'(x)=F'(5)-F'(x^3)$$ Therefore, $$F'(x)=(cos^2(5)-5e^5)-(cos^2(x^3)-x^3e^{x^3})$$ Is this correct? Thank you for any help!

I need to numerically compute integrals such as this (some parameters omitted for simplicity): $$ \int_{0}^{\infty} e^{-x^2} I_{0}(x) K_{0}(x) \mathrm{d}x $$ where $I_{0}$ and $K_{0}$ denote the modified Bessel functions of the first and second kind, respectively. As $x \to \infty$, we have $K_{0} \to 0$ and $I_{0} \to \infty$, but the integrand goes to […]

Wolfram Alpha indicates the following solution form:- $$ \int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx = (1/A)\left( x – \frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} + constant\right)^{2\pi}_0 $$ My thinking is that substituting for $x$ by $2 \pi$ and by $0$ the corresponding values of $\tan(x/2)$ are the same, namely $\tan(\pi) = \tan(0) = […]

Same to the tag calculate $\int_0^{\pi}\frac {x}{1+\cos^2x}dx$. Have no ideas on that. Any suggestion? Many thanks

Can you please help me solve this double integral? $$ \iint_D\ln(\sin(u-7v))\,du\,dv $$ where $$D:=\left\{(u,v)\;:\; 0\le u \le \pi \;,\; 0\le v \le \frac u7\right\} .$$ I know it’s not possible to solve $\int \ln(\sin(x))$ but definite integral is?

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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