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Question : 76 of 160
Marks:
+1,
-0
Solution:
Let
I=∫‌ Putting
x=a‌sec‌θ⇒dx=a‌sec‌θ⋅tan‌θ‌d‌θ I=∫‌| a‌sec‌θ⋅tan‌θ⋅dθ |
| [a2(sec2θ−1)]‌ |
=∫‌×‌| sec‌θ⋅tan‌θ |
| tan3θ2 |
dθ =‌‌∫‌dθ=‌‌∫‌dθ =‌‌∫cot‌θ⋅cosecθdθ=−‌cosec2θ+C =−‌+C (∵sec‌θ=‌⇒cos‌θ=‌∴sin‌θ=√1−‌ ⇒∴cos‌θ=√‌=‌)
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