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Question : 74 of 160
Marks:
+1,
-0
Solution:
∫cos‌√x‌d‌x Let
I=∫cos‌√x‌d‌x Let
√x=t x=t2 Differentiating w.r.t '
x '
l=2t dx=2tdt I=∫cos‌t.2tdt I=2‌∫t‌cos‌t‌d‌t On applying by-part method
I=2[t‌∫cos‌t‌d‌t−∫{t‌∫cos‌t‌d‌t}dt] I=2[t‌sin‌t+cos‌t]+c I=2√x‌sin‌√x+2‌cos‌√x+c
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