Evaluate the indefinite integral of the function 1-sin^2x .

Expert Answers

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We have to find the integral of 1 - (sin x)^2.

1- (sin x)^2 = (cos x)^2

(cos x)^2 = (1/2)(1 + cos 2x)

So, Int[(cos x)^2 dx]

=> Int[(1/2)(1 + cos 2x) dx]

=> (1/2)Int[1 dx] + (1/2)[Int(cos 2x dx)]

=> (1/2)x + (1/2)(sin 2x] + C

=> x/2 +(1/4)sin 2x + C

=> x/2 + (1/4) 2 sin x cos x + C

=> x/2 + (1/2) sin x cos x + C

Therefore the required integral is x/2 + (1/2) sin x cos x + C

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