wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If π20xnsinxdx=(34)(π28). Find n

Open in App
Solution

Consider L.H.S
In=π20xnsinxdx
Evaluating by integration by parts we get
In=xnπ20sinxdx(π20sinxdx)(ddxxn)dxIn=[xncosx]π20+π20nxn1cosxdxIn=[(π2)ncosπ20ncos0]+nπ20xn1cosxdxIn=nπ20xn1cosxdxIn=n[xn1cosxdx(ncosxdxddx(xn1))dx]In=n[(xn1sinx)π20n(n1)π20xn2sinxdx]In=n[(π2)n1sinπ20](n1)nπ20xn2sinxdxIn=n[π2]n1n(n1)In2
Now put n=3
I3=3(π2)26I1=3π246I1
Now to find I1
I1=π20xsinxdx=xπ20sinxdxπ20sinxdx=xcosxπ20+sinxπ20=(π2cosπ2+0.cos0)+(sinπ2sin0)=1I3=3π246=34(π2243)=34(π28)
So comparing this result with out given equation we can say that n=3

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Logarithmic Differentiation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon