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Question : 76 of 160
Marks:
+1,
-0
Solution:
I=sin−1(sin‌x)dx =sin−1(sin‌x)dx+sin−1(sin‌x)dx+sin−1(sin‌x)dx ‌=x‌dx+(π−x)‌dx+(−2π+x)‌dx =[]−π∕2π∕2+[πx−]π∕23π∕2+[−2πx+]3π∕22π =0+‌−‌−+‌−4π2+2π2+‌‌+3π2−‌ =π2+π2−=2π2−=−
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