If I_n = int_0^{pi / 4} tan^n theta , dtheta | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given, $I_n = \int_0^{\pi / 4} 	an^n 	heta \, d	heta$. We need to find $I_{n-1} + I_{n+1}$. $I_{n-1} + I_{n+1} = \int_0^{\pi / 4} 	an^{n-1}

Vedclass · Accepted Answer

$\frac{1}{n}$

Vedclass · Answer

(C) Given, $I_n = \int_0^{\pi / 4} 	an^n 	heta \, d	heta$. We need to find $I_{n-1} + I_{n+1}$. $I_{n-1} + I_{n+1} = \int_0^{\pi / 4} 	an^{n-1} 	heta \, d	heta + \int_0^{\pi / 4} 	an^{n+1} 	heta \, d	heta$. Factor out $	an^{n-1} 	heta$: $I_{n-1} + I_{n+1} = \int_0^{\pi / 4} 	an^{n-1} 	heta (1 + 	an^2 	heta) \, d	heta$. Since $1 + 	an^2 	heta = \sec^2 	heta$, we have: $I_{n-1} + I_{n+1} = \int_0^{\pi / 4} 	an^{n-1} 	heta \sec^2 	heta \, d	heta$. Let $u = 	an 	heta$, then $du = \sec^2 	heta \, d	heta$. When $	heta = 0$, $u = 0$. When $	heta = \pi / 4$, $u = 1$. Substituting these into the integral: $I_{n-1} + I_{n+1} = \int_0^1 u^{n-1} \, du = \left[ \frac{u^n}{n} ight]_0^1 = \frac{1}{n} - 0 = \frac{1}{n}$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

If $I_n = \int_0^{\pi / 4} \tan^n \theta \, d\theta$ for $n = 1, 2, 3, \ldots$, then $I_{n-1} + I_{n+1}$ is equal to

Similar Questions

An increase in $CO_2$ concentration in the atmosphere will result in

Farmers in a particular region were concerned that premature yellowing of leaves of a pulse crop might cause a decrease in the yield. Which treatment could be most beneficial to obtain maximum seed yield?

The distribution of a random variable $X$ is given below:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X = x)$ $\frac{1}{10}$ $k$ $\frac{1}{5}$ $2k$ $\frac{3}{10}$ $k$

The value of $k$ is:

Which of the following oxides cannot act as a reducing agent?

Vedclass Test Series

Exam Paper Generator

Online Exam Module