Assertion (A): If I_n = int cot^n x , dx,then | vedclass

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

Question

(C) Consider the integral $I_n = \int \cot^n x \, dx$.We can write this as $I_n = \int \cot^{n-2} x \cdot \cot^2 x \, dx$.Using the identity $\cot^2 x

Vedclass · Accepted Answer

$A$ is true,$R$ is false

Vedclass · Answer

(C) Consider the integral $I_n = \int \cot^n x \, dx$.We can write this as $I_n = \int \cot^{n-2} x \cdot \cot^2 x \, dx$.Using the identity $\cot^2 x = \csc^2 x - 1$,we get:$I_n = \int \cot^{n-2} x (\csc^2 x - 1) \, dx = \int \cot^{n-2} x \csc^2 x \, dx - \int \cot^{n-2} x \, dx$.For the first part,let $u = \cot x$,then $du = -\csc^2 x \, dx$,so $\csc^2 x \, dx = -du$.Thus,$\int \cot^{n-2} x \csc^2 x \, dx = -\int u^{n-2} \, du = -\frac{u^{n-1}}{n-1} = \frac{-\cot^{n-1} x}{n-1}$.Therefore,the reduction formula is $I_n = \frac{-\cot^{n-1} x}{n-1} - I_{n-2}$.Comparing this with Reason $(R)$,the formula in $(R)$ has $n$ in the denominator instead of $n-1$,so $(R)$ is false.For Assertion $(A)$,using the correct formula for $n=6$:$I_6 = \frac{-\cot^5 x}{5} - I_4 \implies I_6 + I_4 = \frac{-\cot^5 x}{5}$.Thus,Assertion $(A)$ is true and Reason $(R)$ is false.

Assertion $(A)$: If $I_n = \int \cot^n x \, dx$,then $I_6 + I_4 = \frac{-\cot^5 x}{5}$.
Reason $(R)$: $\int \cot^n x \, dx = \frac{-\cot^{n-1} x}{n-1} - \int \cot^{n-2} x \, dx$.

Similar Questions

If $I_n = \int \frac{1}{(x^2+1)^n} dx$, then $2n I_{n+1} - (2n-1) I_n = $

$\int \sin ^{-1}\left(\sqrt{\frac{x-a}{x}}\right) d x=$

Integrate the function: $\frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}}$

$\int \frac{dx}{1-\cos x-\sin x}$ is equal to

If $\int {\frac{{dx}}{{{x^3}{{\left( {1 + {x^6}} \right)}^{2/3}}}} = xf\left( x \right){{\left( {1 + {x^6}} \right)}^{\frac{1}{3}}} + C} $ where $C$ is a constant of integration,then the function $f(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion $(A)$: If $I_n = \int \cot^n x \, dx$,then $I_6 + I_4 = \frac{-\cot^5 x}{5}$.Reason $(R)$: $\int \cot^n x \, dx = \frac{-\cot^{n-1} x}{n-1} - \int \cot^{n-2} x \, dx$.