Introduction to Weierstrass Substitution

Practice problems

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[PXO]

Well I did pretty much the same way but the initial integral can be simplified using identities:
Cosx= 1-tan^2(x/2) /1+tan^2(x/2)
And sinx= 2tan(x/2) /1+tan^(x/2)
Taking the lcm and taking 1/2 common,
We get 1/2 int(sec^2(x/2) /1+2tan(x/2)) dx

Now, the t=tan(x/2) sub is clearly the easiest way to solve it, anyone can tell that from intuition.. It's not an out of the blue substitution as the derivative of tanx is sec^2(x)

Simplifying, we get a much simpler integral : int(dt/(1+2t)) and the answer is 1/2 ln(1+2tan(x/2))

[mohammadw.alomari]

Another Solution: Take 2+2cos(x), from the denominator and set u= sin(x)/(2+2cos(x)), the result becomes very obvious once you find dx, which is indeed dx= (2+2cos(x)) du. Hence, the integral becomes int du/(1+u)= ln |1+u| + C

[mathsplus]

Very interesting problem. Thanks Sir🙏🙏🙏✍💯

[d.h.y]

I really liked it!!! Thank you for letting know this!😃😃