Question

integration of (1/sin^²x)​

Answer 1

Step-by-step explanation:

i dont know whether its right

Answer 2

Answer:

y=∫(√1+sin2x)dx

But 1=sin²x+cos²x

hence

y=∫√(sin²x+cos²x+sin2x)dx

y=∫√(sin²x+cos²x+2sinxcosx)dx beause sin2x=2sinxcosx

y=∫√(sinx+cosx)²dx beause x²+y²+2xy=(x+y)²

y=∫(sinx+cosx)dx

y=-cosx+sinx+c

follow for same answer