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Question

Question: \[\log(a^{2} + b^{2}\sin^{2}x) + c\]...

log(a2+b2sin2x)+c\log(a^{2} + b^{2}\sin^{2}x) + c

A

b2log(a2+b2sin2x)+cb^{2}\log(a^{2} + b^{2}\sin^{2}x) + c

B

1x1+logx6mudx=\int_{}^{}{\frac{1}{x\sqrt{1 + \log x}}\mspace{6mu} dx =}

C

23(1+logx)3/2+c\frac{2}{3}(1 + \log x)^{3/2} + c

D

(1+logx)3/2+c(1 + \log x)^{3/2} + c

Answer

b2log(a2+b2sin2x)+cb^{2}\log(a^{2} + b^{2}\sin^{2}x) + c

Explanation

Solution

(1x)11+c(1 - x)^{- 1} - 1 + c

Put (cosxsinx)6mudx=2sin(x+α)+c\int_{}^{}{(\cos x - \sin x) ⥂ \mspace{6mu} dx = \sqrt{2}\sin(x + \alpha) + c} then it reduces to

α=\alpha =

π3\frac{\pi}{3}

π3- \frac{\pi}{3}

π4\frac{\pi}{4}, π4- \frac{\pi}{4}constant).