Question

If $\int e^{\sin x}\left[\frac{x{\cos }^{3}x-\sin x}{{\cos }^{2}x}\right]dx=e^{\sin x}f\left(x\right)+c$ where $c$ is constant of integration,then $f\left(x\right)=$

• A. $\sec x-x$
• B. $x-\sec x$
• C. $\tan x-x$
• D. $x-\tan x$

Answer 1

Answer: (B)

$x-\sec x$

$\int e^{sinx}\left[\frac{xco{s}^{3}x-sinx}{co{s}^{2}x}\right]$

$=\int e^{sinx}xcosxdx-\int e^{sinx}secxtanxdx+c_1$

$=I_1-I_2$

$I_1=\int e^{sinx}xcosxdx$

$=x\int e^{sinx}cosxdx-\int (\int e^{sinx}cosxdx)\left(\frac{d}{dx}\right(x\left)\right)dx$

$=x{e}^{sinx}-\int e^{sinx}dx+C_{1}dx$

$I_2=\int e^{sinx}secxtanxdx$

$=e^{sinx}\int secxtanxdx-\int (\int secxtanxdx)\left(\frac{d}{dx}\right(e^{sinx}\left)\right)dx$

$=e^{sinx}secx-\int \left(secx\right)\left(e^{sinx}cosx\right)dx$

$=e^{sinx}secx-\int e^{sinx}dx+C_2$

$\therefore I_1-I_2=(x{e}^{sinx}-\int e^{sinx}dx+C_1)-(e^{sinx}secx-\int e^{sinx}dx+C_2)$

$=x{e}^{sinx}-e^{sinx}secx+c$

$=x{e}^{sinx}(x-secx)+c$

$⟹f\left(x\right)=x-secx$

So, the answer is option (B).