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Question: If\(- \frac{1}{3}(1 - \tan^{2}x)^{3/2} + c\), then....

If−13(1−tan⁡2x)3/2+c- \frac{1}{3}(1 - \tan^{2}x)^{3/2} + c, then.

A

−23(1−tan⁡2x)2/3+c- \frac{2}{3}(1 - \tan^{2}x)^{2/3} + c

B

∫sin⁡2xsin⁡5xsin⁡3x 6mudx=\int_{}^{}\frac{\sin 2x}{\sin 5x\sin 3x}\mspace{6mu} dx =

C

logsin3x−logsin5x+c{logsin}3x - {logsin}5x + cany constant

D

13logsin3x+15logsin5x+c\frac{1}{3}{logsin}3x + \frac{1}{5}{logsin}5x + cany constant

Answer

13logsin3x+15logsin5x+c\frac{1}{3}{logsin}3x + \frac{1}{5}{logsin}5x + cany constant

Explanation

Solution

log⁡(1+x2)+c\log(1 + x^{2}) + c

log⁡etan⁡−1x+c\log e^{\tan^{- 1}x} + c

etan⁡−1x+ce^{\tan^{- 1}x} + c

tan⁡−1etan⁡−1x+c\tan^{- 1}e^{\tan^{- 1}x} + c

∫1x(log⁡x)2 6mudx=\int_{}^{}\frac{1}{x(\log x)^{2}}\mspace{6mu} dx = is any constant and 1log⁡x+c\frac{1}{\log x} + c.