Solveeit Logo
Login

Question

Question: \(\int_{}^{}\frac{\sin^{3}xdx}{(\cos^{4}x + 3\cos^{2}x + 1)\tan^{- 1}(\sec x + \cos x)}\) =...

∫sin⁡3xdx(cos⁡4x+3cos⁡2x+1)tan⁡−1(sec⁡x+cos⁡x)\int_{}^{}\frac{\sin^{3}xdx}{(\cos^{4}x + 3\cos^{2}x + 1)\tan^{- 1}(\sec x + \cos x)} =

A

tan–1 (sec x + cos x) + c

B

log tan–1 (sec x + cos x) + c

C

1/(sec x + cos x) + c

D

None of these

Answer

log tan–1 (sec x + cos x) + c

Explanation

Solution

Put tan–1 (sec x + cos x) = t

11+(sec⁡x+cos⁡x)2\frac { 1 } { 1 + ( \sec x + \cos x ) ^ { 2 } } (sec x tan x – sin x) dx = dt

sin x(1cos⁡2x−1)\left( \frac{1}{\cos^{2}x} - 1 \right) dx = dt

cos⁡2xsin⁡x(1−cos⁡2x)cos⁡2xcos⁡2x+1+cos⁡4x+2cos⁡2x\frac { \cos ^ { 2 } x \sin x \frac { \left( 1 - \cos ^ { 2 } x \right) } { \cos ^ { 2 } x } } { \cos ^ { 2 } x + 1 + \cos ^ { 4 } x + 2 \cos ^ { 2 } x } dx = dt

dx = dt

I = ∫dtt\int \frac { \mathrm { dt } } { \mathrm { t } } ̃ log | t | + c

̃ log |tan–1 (sec x + cos x)| + c