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Question : 75 of 160
Marks:
+1,
-0
Solution:
‌‌ (a) Let ‌I=∫etan2θsin‌2θ‌tan‌θ‌d‌θ‌=∫etan2θ×sin‌2θ‌| tan‌θ‌ ‌s‌e‌c2‌θ |
| (1+tan2θ) |
dθ =‌‌∫2etan2θ×‌×‌| tan‌θ‌ ‌s‌e‌c2‌θ |
| (1+tan2θ) |
dθ Let
tan2θ=t⇒2‌tan‌θ‌ ‌s‌e‌c2‌θ‌d‌θ=dtWhen
θ=0, then
t=0and when
θ=‌, then
t=1 ∴I‌=‌‌et×‌‌dt=‌‌et[‌]‌dt‌=‌‌⋅et[‌+⋅(‌)]‌dt =‌[‌]01‌‌[∵∫ex(f(x)+f′(x)‌dx)=exf(x)+C] =‌[‌−1]
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