Assertion (A): if I_n=∫cot^nxdx, then I_6+I_4=(-cot^5x)/5 Reason (R): ∫cot^nxdx=(-cot^(n-1)x)/n-∫cot^(n-2)xdx

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) A is true, R is false. I_n=∫cot^nxdx =∫cot^(n-2)xcot^2xdx =∫cot^(n-2)xcosec^2xdx-∫cot^(n-2)xdx =I-I_(n-2) Here, I=∫cot^(n-2)xcosec^2xdx Let cotx=t; cosec^2xdx=-dt i.e. I=-∫t^(n-2)dt=t^(n-1)1-n=(cot^(n-1)x)/(1-n) I.e. I_n=(cot^(n-1)x)/(1-n)-I_(n-2) ⇒I_6+I_4=-(cot^5x)/5 Hence, assertion is true but reason is false.

Assertion (A): if $I_{n} = \int \cot^{n} x d x$ , then $I_{6} + I_{4} = \frac{- \cot^{5} x}{5}$
Reason (R): $\int \cot^{n} x d x = \frac{- \cot^{n - 1} x}{n} - \int \cot^{n - 2} x d x$

Solution

Relevant Questions

Assertion (A): if In=∫cotnxdx, then I6+I4=-cot5x5Reason (R): ∫cotnxdx=-cotn-1xn-∫cotn-2xdx

Solution

Relevant Questions

Assertion (A): if $I_{n} = \int \cot^{n} x d x$ , then $I_{6} + I_{4} = \frac{- \cot^{5} x}{5}$
Reason (R): $\int \cot^{n} x d x = \frac{- \cot^{n - 1} x}{n} - \int \cot^{n - 2} x d x$