Reduction Formulae for Definite Integration

If $I\left(m,n\right)=\underset{0}{\overset{1}{\int }}{t}^m{\left(1+t\right)}^{n}dt$, then the expression for $I\left(m,n\right)$ in terms of $I\left(m+1,n-1\right)$ where $m, n\in N$ is

If $I_n={\int }_0^1{\left(1+x^2\right)}^{n}dx$, then the value of $13I_6-12I_5$ is

Evaluate ${\int }_0^{\infty }{x}^7{\mathrm{e}}^{-x}dx$.

Evaluate ${\int }_0^{\infty }{e}^{-h^2{x}^2}dx$.

Let $I_n={\int }_0^1{x}^n{\left(1-x\right)}^{\frac{1}{2}} dx,$ for $n⩾0.$ Show that, for $n⩾1, \left(3+2n\right)I_n=2n{I}_{n-1}$. Hence find the exact value of $I_3.$

Given that $l_n={\int }_1^2{\left(\ln \left(x\right)\right)}^n dx$. Show that $I_n=2{\left(\ln \left(2\right)\right)}^n-n{I}_{n-1}$ for $n⩾1.$ Hence, determine the value of $I_5,$ giving your answer correct to $4$ decimal places.

Show that the reduction formula for $I_n={\int }_0^1{x}^n{e}^{3x} dx$ is $I_n=\frac{1}{3}{e}^3-\frac{n}{3}{I}_{n-1}$ for $n⩾1.$ Hence, find $I_3,$ giving your answer in terms of $e$.

If $I_n= \underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^n\theta d\theta$, then the value of $I_n+I_{n-2}$ is equal to

Cotyledons are also called-

Let $I_n={\int }_0^{\frac{\pi }{2}}{x}^n\cos xdx$ , where $n$ is a non-negative integer. Then $\sum _{n=2}^{\infty }\left(\frac{I_n}{n!}+\frac{I_{n-2}}{\left(n-2\right)!}\right)$ equals-

If $I_n=\underset{0}{\overset{\pi /4}{\int }}{\tan }^n\left(x\right)dx$ then $I_2+I_4,I_3+I_5,I_4+I_6,...$ are in ?

If $I_n=\int {\tan }^{n}xdx$ for $n\ge 2$, then $I_n+I_{n-2}$ is equal to

If $I_n={\int }_{-\pi }^{\pi }\frac{\sin nx}{\sin \mathit{x}}dx,n\in N$, then which of the following options are correct ?

${\int }_0^{\infty }{\text{x}}^{2\text{n}+1}·{\text{e}}^{-{\text{x}}^2}\text{dx}$ is equal to $\left(\text{n }ϵ\text{ N}\right)$