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Question

Question: \[x^{2} = t\]...

x2=tx^{2} = t

A

∫x6tan⁡−1x3 6mudx\int_{}^{}{x^{6}\tan^{- 1}x^{3}}\mspace{6mu} dx

B

∫tan⁡−1(2x1−x2) 6mudx\int_{}^{}{\tan^{- 1}\left( \frac{2x}{1 - x^{2}} \right)\mspace{6mu} dx}

C

∫x3cos⁡x2 6mudx\int_{}^{}{x^{3}\cos x^{2}\mspace{6mu} dx}

D

∫tan⁡xsec⁡2x1−tan⁡2x⥂ 6mudx=\int_{}^{}{\tan x}\sec^{2}x\sqrt{1 - \tan^{2}x} ⥂ \mspace{6mu} dx =

Answer

∫tan⁡xsec⁡2x1−tan⁡2x⥂ 6mudx=\int_{}^{}{\tan x}\sec^{2}x\sqrt{1 - \tan^{2}x} ⥂ \mspace{6mu} dx =

Explanation

Solution

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

=∫sec⁡2xdx+∫tan⁡x⋅sec⁡xdx= \int \sec ^ { 2 } x d x + \int \tan x \cdot \sec x d x ∫etan⁡−1x1+x2dx=\int_{}^{}{\frac{e^{\tan^{- 1}x}}{1 + x^{2}}dx =}.