If I_n = int_0^{pi / 4} tan^n x , dx, then I_ | vedclass

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

Question

(C) Given $I_n = \int_0^{\pi / 4} 	an^n x \, dx$. Consider the sum $I_r + I_{r+2} = \int_0^{\pi / 4} 	an^r x \, dx + \int_0^{\pi / 4} 	an^{r+2} x \

Vedclass · Accepted Answer

harmonic progression

Vedclass · Answer

(C) Given $I_n = \int_0^{\pi / 4} 	an^n x \, dx$. Consider the sum $I_r + I_{r+2} = \int_0^{\pi / 4} 	an^r x \, dx + \int_0^{\pi / 4} 	an^{r+2} x \, dx$. $I_r + I_{r+2} = \int_0^{\pi / 4} 	an^r x (1 + 	an^2 x) \, dx$. Since $1 + 	an^2 x = \sec^2 x$, we have $I_r + I_{r+2} = \int_0^{\pi / 4} 	an^r x \sec^2 x \, dx$. Let $t = 	an x$, then $dt = \sec^2 x \, dx$. When $x=0, t=0$ and when $x=\pi/4, t=1$. Thus, $I_r + I_{r+2} = \int_0^1 t^r \, dt = \left[ \frac{t^{r+1}}{r+1} ight]_0^1 = \frac{1}{r+1}$. For $r=2$, $I_2 + I_4 = \frac{1}{3}$. For $r=3$, $I_3 + I_5 = \frac{1}{4}$. For $r=4$, $I_4 + I_6 = \frac{1}{5}$. The sequence is $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots$, which are in harmonic progression $(HP)$ because their reciprocals $3, 4, 5, \ldots$ are in arithmetic progression.

If $I_n = \int_0^{\pi / 4} \tan^n x \, dx$, then $I_2+I_4, I_3+I_5, I_4+I_6, \ldots$ are in

Similar Questions

Hormones produce their effect on target tissue by binding to specific proteins called as

Which of the following is used as an antibiotic?

Sunken stomata are found in the leaves of

Under which one of the following conditions do real gases approach the ideal gas behaviour?

If $f:R \to R$ and $g:R \to R$ are defined by $f(x) = 2x + 3$ and $g(x) = x^2 + 7$,then the values of $x$ such that $g(f(x)) = 8$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module