Is it possible to integrate this?

Question

$$\int\frac{e^{\sin x }(x \cos ^3 x- \sin x)}{\cos ^2 x}dx$$. What is to be substituted for the integration ?

Answer 1

Let $$I_1=\int e^{\sin x}x\cos xdx$$

Do this by parts with $u=x$ and get $$I_1=xe^{\sin x}-\int e^{\sin x}dx$$

Now let $$I_2=\int e^{\sin x}\sec x\tan xdx$$ so that the required integral is $I_1-I_2$

Do this second integral also by parts with $u=e^{\sin x}$ and the unintegrable term will cancel to give $$xe^{\sin x}-e^{\sin x}\sec x+c$$

Answer 2

$I = \int\frac{e^{\sin x }(x \cos ^3 x- \sin x)}{\cos ^2 x}dx$

$I = \int e^{\sin(x)}(x\cos(x)-\sec(x)\tan(x))\,dx$

$I = \int e^{\sin(x)}(x\cos(x)-1+1-\sec(x)\tan(x))\,dx$

$I = \int e^{\sin(x)}(1-\sec(x)\tan(x))+e^{\sin(x)}(x\cos(x)-1)\,dx$

$I =\int e^{\sin(x)}(1-\sec(x)\tan(x))+e^{\sin(x)}\cos(x)(x-\sec(x))\,dx$

Notice that $\frac{d}{dx}( e^{\sin(x)}(x-\sec(x))) =e^{\sin(x)}(1-\sec(x)\tan(x))+e^{\sin(x)}\cos(x)(x-\sec(x))$

Therefore ;

$I = e^{\sin(x)}(1-\sec(x)\tan(x))+C$