For any integer n geq 2, let I_n = int tan^n | vedclass

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

Question

(A) Given $I_n = \int 	an^n x \, dx$. We can write this as: $I_n = \int 	an^{n-2} x \cdot 	an^2 x \, dx$ $I_n = \int 	an^{n-2} x (\sec^2 x - 1) \,

Vedclass · Accepted Answer

$(n-1, 1)$

Vedclass · Answer

(A) Given $I_n = \int 	an^n x \, dx$. We can write this as: $I_n = \int 	an^{n-2} x \cdot 	an^2 x \, dx$ $I_n = \int 	an^{n-2} x (\sec^2 x - 1) \, dx$ $I_n = \int 	an^{n-2} x \sec^2 x \, dx - \int 	an^{n-2} x \, dx$ Using the substitution $u = 	an x$, $du = \sec^2 x \, dx$, the first integral becomes $\int u^{n-2} \, du = \frac{u^{n-1}}{n-1} = \frac{	an^{n-1} x}{n-1}$. Thus, $I_n = \frac{	an^{n-1} x}{n-1} - I_{n-2}$. Comparing this with the given form $I_n = \frac{1}{a} 	an^{n-1} x - b I_{n-2}$, we get: $\frac{1}{a} = \frac{1}{n-1} \implies a = n-1$ $b = 1$ Therefore, the ordered pair $(a, b) = (n-1, 1)$.

For any integer $n \geq 2$, let $I_n = \int \tan^n x \, dx$. If $I_n = \frac{1}{a} \tan^{n-1} x - b I_{n-2}$ for $n \geq 2$, then the ordered pair $(a, b)$ is equal to

Similar Questions

If $\int {\frac{{\cos 4x + 1}}{{\cot x - \tan x}}} dx = k\,\cos 4x + c$,then:

If $\int \frac{d \theta}{\cos ^2 \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log _{e}|f(\theta)|+c$ (where $c$ is a constant of integration),then the ordered pair $(\lambda, |f(\theta)|)$ is equal to

$\int \sqrt{x^2 - 8x + 7} \, dx = $

$\int \frac{dx}{\sin^6 x + \cos^6 x} = $

If $\int \frac{\cos x \, dx}{\sin ^{3} x \left(1+\sin ^{6} x\right)^{2 / 3}} = f(x) \left(1+\sin ^{6} x\right)^{1 / \lambda} + c$ where $c$ is a constant of integration,then $\lambda f\left(\frac{\pi}{3}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module