Evaluate-the-following-integral-I-pi-4-pi-2-cos2x-sin2x-ln-cosx-sinx-dx-

Question Number 1350 by 112358 last updated on 24/Jul/15

E v a l u a t e t h e f o l l o w i n g i n t e g r a l :

I = \int_{π / 4}^{π / 2} (c o s 2 x + s i n 2 x) l n (c o s x + s i n x) d x

Commented by prakash jain last updated on 25/Jul/15

\int \sin 2 x \ln (\cos x + \sin x) d x

= \ln (\cos x + \sin x) (- \frac{\cos 2 x}{2}) + \int \frac{\cos 2 x}{2} \cdot \frac{\cos x - \sin x}{\cos x + \sin x} d x

\int \frac{\cos 2 x}{2} \cdot \frac{\cos x - \sin x}{\cos x + \sin x} d x = \frac{1}{2} \int (1 - \sin 2 x) d x

= \frac{x}{2} + \frac{\cos 2 x}{4}

From above

\int \sin 2 x \ln (\cos x + \sin x) d x =

= \frac{x}{2} + \frac{\cos 2 x}{4} - \frac{\cos 2 x}{2} \ln (\cos x + \sin x)

Commented by prakash jain last updated on 25/Jul/15

\int \cos 2 x \ln (\cos x + \sin x) d x

= \ln (\cos x + \sin x) \frac{\sin 2 x}{2} - \int \frac{\sin 2 x}{2} \cdot \frac{\cos x - \sin x}{\cos x + \sin x} d x

\int \frac{\sin 2 x}{2} \cdot \frac{\cos x - \sin x}{\cos x + \sin x} d x =

\frac{1}{2} [\sin x \cos x - \ln (\cos x + \sin x)

Commented by 112358 last updated on 25/Jul/15

T h a n k s .

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