B-1-sin-6x-1-sin-6x-dx-

Question Number 134641 by benjo_mathlover last updated on 06/Mar/21

B = \int \frac{1 - \sin 6 x}{1 + \sin 6 x} d x

Answered by Dwaipayan Shikari last updated on 06/Mar/21

\int \frac{2}{1 + s i n 6 x} - 1 d x 6 x = u

= \frac{1}{3} \int \frac{1}{1 + s i n u} d u - x

= \frac{2}{3} \int \frac{1}{1 + \frac{2 t}{1 + t^{2}}} . \frac{1}{1 + t^{2}} d t - x t a n \frac{u}{2} = t

= \frac{2}{3} \int \frac{1}{{(1 + t)}^{2}} d t - x = - \frac{2}{3 (1 + t)} - x + C = - \frac{2}{3 (1 + t a n 3 x)} - x + C

Answered by Olaf last updated on 06/Mar/21

By using \sin 6 x = \frac{2 \tan (3 x)}{1 + \tan^{2} (3 x)}

and integrating by parts :

B = - \frac{2}{3} . \frac{1}{1 + \tan (3 x)} - x + C

Answered by EDWIN88 last updated on 06/Mar/21

B = \int \frac{1 - \sin 6 x}{1 + \sin 6 x} dx = \int \tan^{2} (\frac{π}{4} - 3 x) dx

B = \int [\sec^{2} (\frac{π}{4} - 3 x) - 1] dx

B = - \frac{1}{3} \tan (\frac{π}{4} - 3 x) - x + c

Answered by Ar Brandon last updated on 06/Mar/21

B = \int \frac{1 - \sin 6 x}{1 + \sin 6 x} dx = \int \frac{{(1 - \sin 6 x)}^{2}}{\cos^{2} 6 x} dx

= \int \frac{1 - 2 \sin 6 x + \sin^{2} 6 x}{\cos^{2} 6 x} dx = \int \frac{2 - 2 \sin 6 x - \cos^{2} 6 x}{\cos^{2} 6 x}

= \frac{\tan 6 x}{3} - \frac{1}{3 \cos 6 x} - x + C