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How do you evaluate the definite integral #intcosx/(1+sin^2x)dx# from #[0,1]#?

Answer 1

Christopher Agresta

The answer is #=0.6995#

We need

#int(dx)/(1+x^2)=arctan(x)+ C#

First determine the definite integral by substitution

Let #u=sinx#, #=>#, #du=cosxdx#

Therefore,

#int(cosxdx)/(1+sin^2x)=int(du)/(u^2+1)#

#=arctan(u)#

#=arctan(sinx)+C#

The definite integral is

#int_0^1(cosxdx)/(1+sin^2x)#

#=[arctan(sinx)]_0^1#

#=(arctansin1)-(0)#

#=0.6995#

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Answer 2

Christopher Allers

To evaluate the definite integral ∫₀¹ cos(x)/(1 + sin²(x)) dx, you can use the trigonometric substitution method. Let u = sin(x), then du = cos(x) dx. After substitution, the integral becomes ∫₀¹ du/(1 + u²). This integral can be evaluated using the arctangent function.

So, the integral becomes arctan(u) evaluated from 0 to 1. Substituting back u = sin(x), you get arctan(sin(x)) evaluated from 0 to 1.

Evaluate arctan(sin(1)) - arctan(sin(0)) to find the value of the definite integral.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.