Question

$\int \frac{1}{\sin x\cos x({\tan }^{9}x+1)}dx=$

• A. $\log |\frac{{\cos }^{9}x}{{\sin }^{9}x+{\cos }^{9}x}|+c$
• B. $\frac{1}{9}\log |\frac{{\sin }^{9}x}{{\sin }^{9}x+{\cos }^{9}x}|+c$
• C. $\frac{1}{9}\log |\frac{{\sin }^{9}x+{\cos }^{9}x}{{\sin }^{9}x}|+c$
• D. $\frac{1}{9}\log |\frac{{\cos }^{9}x}{{\sin }^{9}x+{\cos }^{9}x}|+c$

Answer 1

The correct option is B $\frac{1}{9}\log |\frac{{\sin }^{9}x}{{\sin }^{9}x+{\cos }^{9}x}|+c$

$\int \frac{se{c}^{2}x}{\left(tanx\right)(ta{n}^{9}x+c)}dx$
Let, $ta{n}^x=t$
$se{c}^{2}xdx=dt$
$\frac{1}{9}\int \frac{9t^{8}dt}{t^9(t^9+c)}$
$\frac{1}{9}(\log t^9-\log (t^9+c\left)\right)$
Put back, $t=ta{n}^x$
$\frac{1}{9}\log \left(\frac{si{n}^{9}x}{si{n}^{9}x+co{s}^{9}x}\right)$