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Question : 103 of 120
Marks:
+1,
-0
Solution:
Let
I=∫sin−1[]‌dx ⇒
I=∫sin−1[]‌dx I=∫sin−1[]‌dx Substituting 2x + 2 = 3 tan θ,
⇒2‌dx=3sec2‌θdθ, we get
I=∫sin−1[].sec2‌θdθ ⇒
I=‌∫sin−1(sin‌θ).sec2‌θdθ ⇒
I=‌∫θsec2‌θdθ ⇒I=[θ‌tan‌θ−∫tan‌θdθ]‌‌‌‌‌(integrating‌by‌parts) I=[θ‌tan‌θ−log|sec‌θ|]+c =[tan−1().()−log‌√1+()2]+c =(x+1)tan−1()−‌log()+c
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