Antgral sin radical x dx
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Use the substitution t=x−−√, and then use integration by parts. From the substitution, you should get:
∫sinx−−√dx=2∫tsin(t)dt
Then, use integration by parts. Let u=t and dv=sint. Then du=dt and v=−cost, and we have:
∫tsin(t)dt=−tcost+∫cos(t)dt=sint−tcost=sinx−−√−x−−√cosx−−√
and we have the solution:
∫sinx−−√dx=2(sinx−−√−x−−√cosx−−√)
I hope this clears up the matter for
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