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Question : 75 of 160
Marks:
+1,
-0
Solution:
Consider the given expression
In=∫dx⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) And
In−2=∫dx⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II) Subtract equation (I) and (II),
In−In−2=∫| (sinnx−sin(n−2)x) |
| sinx |
dx =∫dx =2cos(n−1)xdx = This implies,
I6−I4= and I4−I2= Now,
I2=∫dx =∫dx =2∫cosxdx =2sinx+c And,
I6=I4+2 =I2+2+2 =2+2+2sinx+c =+(3sinx−4sin3x)+2sinx+c So
I6=sin5x−sin3x+4sinx+c
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