1
Question
Solve : ∫sin−1x.1x2dx
Open in App
Solution
∫sin−1x1x2dx
=sin−1x−1x−∫1√1−x2(−1x)
=−sin−1xx+∫1x√1−x2dx
=−sin−1xx+ln(∣∣√1−x2−1∣∣)−ln(√1−x2+1)2+C
∫sin−1x1x2dx
=sin−1x−1x−∫1√1−x2(−1x)
=−sin−1xx+∫1x√1−x2dx
=−sin−1xx+ln(∣∣√1−x2−1∣∣)−ln(√1−x2+1)2+C
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
