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Question

Mathematics Question on Trigonometric Identities

∫11+3sin⁡2x+8cos⁡2x dx=\int\frac{1}{1+3\sin^2 x+8 \cos^2 x}\ dx=

A

16tan⁡−1(2tan⁡x3)+C\frac{1}{6}\tan^{-1}(\frac{2\tan x}{3})+C

B

16tan⁡−1(2tan⁡x)+C\frac{1}{6}\tan^{-1}(2\tan x)+C

C

6tan⁡−1(2tan⁡x3)+C6\tan^{-1}(\frac{2\tan x}{3})+C

D

tan⁡−1(2tan⁡x3)+C\tan^{-1}(\frac{2\tan x}{3})+C

Answer

16tan⁡−1(2tan⁡x3)+C\frac{1}{6}\tan^{-1}(\frac{2\tan x}{3})+C

Explanation

Solution

The correct answer is (A) : 16tan⁡−1(2tan⁡x3)+C\frac{1}{6}\tan^{-1}(\frac{2\tan x}{3})+C.