Question

If $I=\int e^{x\sin x+\cos x}\left(\frac{x^4{\cos }^{3}x-x\sin x+\cos x}{x^2{\cos }^{2}x}\right)dx$ then I equals

• A. $I=e^{x\sin x+\cos x}(x-\frac{\sec x}{x})+C$
• B. $I=e^{x\sin x+\cos x}(x\sin x-\frac{\cos x}{x})$
• C. $I=e^{x\sin x+\cos x}(\frac{x}{\tan x}-\frac{\cos x}{x})+C$
• D. $I=x{e}^{x\sin x+\cos x}-\int e^{x\sin x+\cos x}(1-\frac{\cos x-x\sin x}{x^2{\cos }^{2}x})dx$

Answer 1

Answer: (A), (D)

$I=e^{x\sin x+\cos x}(x-\frac{\sec x}{x})+C$
$I=x{e}^{x\sin x+\cos x}-\int e^{x\sin x+\cos x}(1-\frac{\cos x-x\sin x}{x^2{\cos }^{2}x})dx$

$I=\int e^{x\sin x+\cos x}\left(\frac{x^4{\cos }^{3}x-x\sin x+\cos x}{x^2{\cos }^{2}x}\right)dx$
$=\int e^{x\sin x+\cos x}(x^2\cos x+\frac{\cos x-x\sin x}{{\left(x\cos x\right)}^2})dx$
$=I_1+I_2$
Where $I_1=\int \left(e^{x\sin x+\cos x}x\cos x\right)xdx$
$=x{e}^{x\sin x+\cos x}-\int e^{x\sin x+\cos x}dx$
And $I_2=\int e^{x\sin x+\cos x}\frac{(\cos x-x\sin x)}{{\left(x\cos x\right)}^2}dx$
$=e^{x\sin x+\cos x}(-\frac{1}{x\cos x})+\int e^{x\sin x+\cos x}\left(x\cos x\right)\frac{dx}{x\cos x}$
Therefore $I=e^{x\sin x+\cos x}(x-\frac{1}{x\cos x})+c$