Mathematics › Indefinite Integration › Integration using Reduction Formula
Assertion (A): if In=∫cotnxdx, then I6+I4=-cot5x5
Reason (R): ∫cotnxdx=-cotn-1xn-∫cotn-2xdx
In=∫cotnxdx=∫cotn-2xcot2xdx=∫cotn-2xcosec2xdx-∫cotn-2xdx=I-In-2Here, I=∫cotn-2xcosec2xdxLet cotx=t; cosec2xdx=-dti.e. I=-∫tn-2dt=tn-11-n=cotn-1x1-nI.e. In=cotn-1x1-n-In-2⇒I6+I4=-cot5x5Hence, assertion is true but reason is false.
In=∫cotnxdx
=∫cotn-2xcot2xdx
=∫cotn-2xcosec2xdx-∫cotn-2xdx
=I-In-2
Here, I=∫cotn-2xcosec2xdx
Let cotx=t; cosec2xdx=-dt
i.e. I=-∫tn-2dt=tn-11-n=cotn-1x1-n
I.e. In=cotn-1x1-n-In-2
⇒I6+I4=-cot5x5
Hence, assertion is true but reason is false.
Asked in: AP EAMCET 2022 (04 Jul Shift 1)
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