Question
If In=∫tannxdx,then I0+I1+2(I2+...I8)+I9+I10, is equal to
Answer: Option A
:
A
We have In=∫tannxdx=∫tann−2x(sec2x−1)dx=tann−1xn−1−In−2i.e.In−2+In=tann−1xn−1(n≥2)
Thus, we have
I0+I1+2(I2+....+I8)+I9+I10=(I0+I2)+(I1+I3)+(I2+I4)+(I3+I5)+(I4+I6)+(I5+I7)+(I6+I8)+(I7+I9)+(I8+I10).=∑10n=2tann−1xn−1=∑9n=1tannxn
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:
A
We have In=∫tannxdx=∫tann−2x(sec2x−1)dx=tann−1xn−1−In−2i.e.In−2+In=tann−1xn−1(n≥2)
Thus, we have
I0+I1+2(I2+....+I8)+I9+I10=(I0+I2)+(I1+I3)+(I2+I4)+(I3+I5)+(I4+I6)+(I5+I7)+(I6+I8)+(I7+I9)+(I8+I10).=∑10n=2tann−1xn−1=∑9n=1tannxn
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