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Question : 78 of 160
Marks:
+1,
-0
Solution:
‌‌ Let ‌In=tann(‌)‌dx‌In−2=(tan‌)n−2‌dx‌⇒‌‌In+In−2=(tan‌)n−2× sec2‌‌dx‌⇒‌‌In+In−2=2‌tn−2‌dt=‌‌⇒‌‌I14+I12=‌‌⇒‌‌I12+I10=‌‌⇒‌‌I10+I8=‌‌⇒‌‌I8+I6=‌‌⇒‌‌I6+I4=‌‌⇒‌‌I4+I2=‌‌⇒‌‌I2+I0=2‌∴(i)−(ii)+(iii)−(iv)+(v)−(vi)+(vii)‌I14=2[f(x)−‌]‌∴f(n)=‌−‌+‌−‌....(iv)‌f(n)=‌
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