If In=∫sinnxdx, then nIn-(n-1)In-2=?
sinn-1xcosx
cosn-1xsinx
-sinn-1xcosx
-cosn-1xsinx
Explanation for the correct option.
Step 1: Solve In=∫sinnxdx.
In=∫sinnxdx=∫sinn-1x·sinxdx
Integrating by part, we get
In=sinn-1x-cosx-∫n-1sinn-2x·cosx·-cosxdx=-sinn-1x·cosx+∫n-1sinn-2x·cos2xdx=-sinn-1x·cosx+∫n-1sinn-2x·1-sin2xdx=-sinn-1x·cosx+∫n-1sinn-2x-sinnxdx=-sinn-1x·cosx+n-1∫sinn-2xdx-n-1∫sinnxdx=-sinn-1x·cosx+n-1In-2-n-1In....1
Step 2: Find the value of nIn-(n-1)In-2.
From 1, we get
In=-sinn-1x·cosx+n-1In-2-n-1In⇒In-n-1In-2+n-1In=-sinn-1x·cosx⇒n-1+1In-n-1In-2=-sinn-1x·cosx⇒nIn-n-1In-2=-sinn-1x·cosx
Hence, option C is correct.