Evaluate the definite integral of y=1/cos ^2x.

x=0 to x=pi/4

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The definite integral will be evaluated using the Leibniz-Newton formula.

Int f(x)dx = F(b) - F(a), where x = a to x = b

We'll put y = f(x) = 1/(cos x)^2

We'll compute the indefinite integral, first:

Int dx/(cos x)^2 = tan x + C

We'll note the result F(x) = tan x + C

We'll determine F(a), for a = 0:

F(0) = tan 0

F(0) = 0

We'll determine F(b), for b = pi/4:

F(pi/4) = tan pi/4

F(pi/4) = 1

We'll evaluate the definite integral:

Int dx/(cos x)^2 = F(pi/4) - F(0)

Int dx/(cos x)^2 = 1 - 0

**Int dx/(cos x)^2 = 1, from x = 0 to x = pi/4**

To Evaluate the definite integral of y=1/cos ^2x for x=0 to x=pi/4.

Given y = 1/(cos^2x) = (secx)^2 , 1/cosx = secx.

We know that if F(x) = tanx,

then F'(x) = (tanx)' = (secx)^2.

Therefore Int F'(x) dx = Int (secx)^2 dx = tanx.

Therefore Int (secx)^2 dx = F(x) = tanx.

Therefore area under definite int y = 1/(cosx)^2 from x= 0 to x= pi/4 is F(pi/4)-F(0).

F(pi/4) -F(0) = tanpi/4 - tan 0 .

F(pi/4) - F(0) = 1-0.

F(pi/4) - F(0) = 1.

Therefore the area under the curve is 1 sq unit.

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