Question

The integral $\int (3x^{2}tan\frac{1}{x}-xse{c}^2\frac{1}{x})dx$ equals to

• A. $x(tan\frac{1}{x}+sec\frac{1}{x})+c$
• B. $x^{2}tan\frac{1}{x}-sec\frac{1}{x}+c$
• C. $x^{3}tan\frac{1}{x}+c$
• D. $(x^3-1)tan\frac{1}{x}+c$

Answer 1

Answer: (C)

$x^{3}tan\frac{1}{x}+c$

$=\int (3x^{2}tan\frac{1}{x}-xse{c}^2\frac{1}{x})dx$

observe that the product rule of differentiation is as follows

$\frac{d}{dx}\left(f\right(x).g(x\left)\right)=\frac{d}{dx}f\left(x\right).g\left(x\right)+f\left(x\right)\frac{d}{dx}g\left(x\right)$

consider the function $x^{3}tan\frac{1}{x},$ then

$\frac{d}{dx}\left(x^{3}tan\frac{1}{x}\right)=3x^{2}tan\frac{1}{x}+x^{3}se{c}^2\frac{1}{x}.\left(\frac{-1}{x^2}\right)$

$=3x^{2}tan\frac{1}{x}-xse{c}^2\frac{1}{x}$

Thus,

$I=\int (3x^{2}tan\frac{1}{x}-xse{c}^2\frac{1}{x})dx=x^{3}tan\frac{1}{x}+C$