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Question : 77 of 160
Marks:
+1,
-0
Solution:
Let
I=sin‌4θcos3θdθ =sin‌4θ+(1−sin‌2θ)‌cos‌θ‌d‌θ Let
sin‌θ=t, then
cos‌θ‌d‌θ=dt When
θ→0,t→0 and
θ→‌,t→1 I‌=t4(1−t2)‌dt‌=(t4−t6)‌dt‌=[‌−‌]01=‌−‌‌=‌
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