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Evaluate the following integral:
∫√cos2xsinxdx
Plz show me the steps also.
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5 Answers
Hari Shankar
·Dec 24 '11 at 23:03
Hari Shankar
·Dec 25 '11 at 3:54
Nice jagdish.
My approach (its written with some steps with trig manipulation skipped):
∫sinx√cos2x dx=2∫(cosx+sinx)−(cosx−sinx)√(cosx+sinx)(cosx−sinx) dx
∫tan(4π+x)−1√tan(4π+x) dx=∫tany−1√tany dy
Now let √tany=z
Then the integral becomes
∫(z2−1)(1+z4)2z2 dz=∫(z2−1)(1+z4)2z2 dz
∫(z2−1)(1+z4)(z2−1)2−(z4+1) dz=∫1+z4(z2−1)−z2−11 dz
For the 1st one, divide by z2 and use the substitution
z+z1=t
Second integral is well known and easy