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Mathematics Question on Integration

∫2tan⁡x+3sin⁡2x+2cos⁡2xdx=\int\frac{2\tan x+3}{\sin^2x+2\cos^2x}dx=

A

32sin−1sin⁡x2+ln∣sin⁡2x+2∣+C\frac{3}{\sqrt2}sin^{-1}\frac{\sin x}{\sqrt2}+ln|\sin^2x+2|+C

B

32tan−1tan⁡x2+ln∣tan⁡2x+2∣+C\frac{3}{\sqrt2}tan^{-1}\frac{\tan x}{\sqrt2}+ln|\tan^2x+2|+C

C

32tan−1tan⁡x2−ln∣tan⁡2x+2∣+C\frac{3}{\sqrt2}tan^{-1}\frac{\tan x}{\sqrt2}-ln|\tan^2x+2|+C

D

32cos−1cos⁡x2+ln∣sin⁡2x+2∣+C\frac{3}{\sqrt2}cos^{-1}\frac{\cos x}{\sqrt2}+ln|\sin^2x+2|+C

E

32cos−1cos⁡x2−ln∣cos⁡2x+2∣+C\frac{3}{\sqrt2}cos^{-1}\frac{\cos x}{\sqrt2}-ln|\cos^2x+2|+C

Answer

32tan−1tan⁡x2+ln∣tan⁡2x+2∣+C\frac{3}{\sqrt2}tan^{-1}\frac{\tan x}{\sqrt2}+ln|\tan^2x+2|+C

Explanation

Solution

The correct option is (B): 32tan−1tan⁡x2+ln∣tan⁡2x+2∣+C\frac{3}{\sqrt2}tan^{-1}\frac{\tan x}{\sqrt2}+ln|\tan^2x+2|+C