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Choose the correct answers in Exercises 41 to 44.

    Q42.    \int\frac{\cos 2x}{(\sin x + \cos x)^2}dx is equal to

                (A)    \frac{-1}{\sin x + \cos x} + C

                (B)    \log |{\sin x + \cos x} |+ C

                (C)    \log |{\sin x- \cos x} |+ C

                (D)    \frac{1}{(\sin x + \cos x)^2} + C

Answers (1)

best_answer

\\\frac{\cos 2x}{(\sin x + \cos x)^2}\\ =\frac{cos^{2}x-sin^{2}x}{(\sin x + \cos x)^2}\\ =\frac{(\sin x + \cos x)(\cos x-\sin x)}{(\sin x + \cos x)^2} \\=\frac{(\cos x-\sin x)}{(\sin x + \cos x)}                     cos2x=cos2x-sin2x

let sinx+cosx=t,(cosx-sinx)dx=dt

hence the given integral can be written as

\\\int\frac{\cos 2x}{(\sin x + \cos x)^2}dx\\ =\int \frac{dt}{t}\\ =log|t|+c \\=log|cosx+sinx|+c

B is correct

Posted by

Sayak

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