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Question: \[\int_{}^{}{\frac{\sin x\cos x}{a\cos^{2}x + b\sin^{2}x}dx =}\]...

∫sin⁡xcos⁡xacos⁡2x+bsin⁡2xdx=\int_{}^{}{\frac{\sin x\cos x}{a\cos^{2}x + b\sin^{2}x}dx =}

A

12(b−a)log⁡(acos⁡2x+bsin⁡2x)+c\frac{1}{2(b - a)}\log(a\cos^{2}x + b\sin^{2}x) + c

B

1b−alog⁡(acos⁡2x+bsin⁡2x)+c\frac{1}{b - a}\log(a\cos^{2}x + b\sin^{2}x) + c

C

12log⁡(acos⁡2x+bsin⁡2x)+c\frac{1}{2}\log(a\cos^{2}x + b\sin^{2}x) + c

D

None of these

Answer

1b−alog⁡(acos⁡2x+bsin⁡2x)+c\frac{1}{b - a}\log(a\cos^{2}x + b\sin^{2}x) + c

Explanation

Solution

Put

∫1+sin⁡x2 6mudx=\int_{}^{}{\sqrt{1 + \sin\frac{x}{2}}\mspace{6mu} dx =}

Then 14(cos⁡x4−sin⁡x4)+c\frac{1}{4}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right) + c

4(cos⁡x4−sin⁡x4)+c4\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right) + c