Question

solve this question ☺️☺️​

Answer 1

$\textsf{Step-by-step explanation:}$

$\mathsf{Let,\:\big(x-\frac{1}{x\:cosx}\big).e^{x\:sinx+cosx}=z}$

$\to \mathsf{\frac{x^{2}\:cosx-1}{x\:cosx}.e^{x\:sinx+cosx}=z}$

$\therefore \mathsf{\big[\frac{x\:cosx\:(2x\:cosx-x^{2}\:sinx)-(x^{2}\:cosx-1)(cosx-x\:sinx)}{x^{2}\:cos^{2}x}.e^{x\:sinx+cosx}+\frac{x^{2}\:cosx-1}{x\:cosx}.x\:cosx.e^{x\:sinx+cosx}\big]dx=dz}$

$\to \mathsf{\big[\frac{2x^{2}\:cos^{2}x-x^{3}\:sinx\:cosx+x^{3}\:sinx\:cosx-x^{2}\:cos^{2}x+cosx-x\:sinx}{x^{2}\:cos^{2}x}+(x^{2}\:cosx-1)\big].e^{x\:sinx+cosx}\:dx=dz}$

$\to \mathsf{\frac{x^{2}\:cos^{2}x+cosx-x\:sinx+x^{4}\:cos^{3}x-x^{2}\:cos^{2}x}{x^{2}\:cos^{2}x}.e^{x\:sinx+cosx}\:dx=dz}$

$\to \mathsf{e^{x\:sinx+cosx}.\big[\frac{x^{4}\:cos^{3}x-x\:sinx+cosx}{x^{2}\:cos^{2}x}\big]\:dx=dz}$

$\textsf{On integration, we get}$

$\int \mathsf{e^{x\:sinx+cosx}.\big[\frac{x^{4}\:cos^{3}x-x\:sinx+cosx}{x^{2}\:cos^{2}x}\big]\:dx=\int dz}$

$\mathsf{=z+C,}\:\textsf{where C is constant of integration}$

$\mathsf{=\big(x-\frac{1}{x\:cosx}\big).e^{x\:sinx+cosx}+C,}$

$\textsf{which is the required integral.}$

Answer 2

Given:

$\int e^{(x\sin x+\cos x)}\left[\frac{x^4\cos^3x-x\sin x+\cos x}{x^2\cos^2x}\right]dx$

Solution:

Let us assume that:  $(x-\frac{1}{x\cos x})e^{x\sin x+\cos x}=t}$

${\frac{x^{2}\cos x-1}{x\cos x}e^{x\sin x+cos x}=t}$

Use the quotient rule of integration.

${[\frac{x\cos x(2x\cos x-x^{2}\sin x)-(x^{2}\cos x-1)(cos x-x\sin x)}{x^{2}\cos^{2}x}e^{x\sin x+cos x}+\frac{x^{2}\cos x-1}{x\cos x}x\cos x(e^{x\sin x+\cos x}dx=dt}$ $\mathsf{\big[\frac{2x^{2}\:cos^{2}x-x^{3}\:sinx\:cosx+x^{3}\:sinx\:cosx-x^{2}\:cos^{2}x+cosx-x\:sinx}{x^{2}\:cos^{2}x}+(x^{2}\:cosx-1)\big]e^{x\:sinx+cosx}\:dx=dt}\\ \mathsf{\frac{x^{2}\:cos^{2}x+cosx-x\:sinx+x^{4}\:cos^{3}x-x^{2}\:cos^{2}x}{x^{2}\:cos^{2}x}e^{x\:sinx+cosx}\:dx=dt} \\\mathsf{e^{x\:sinx+cosx}\big[\frac{x^{4}\:cos^{3}x-x\:sinx+cosx}{x^{2}\:cos^{2}x}\big]\:dx=dt}$

Integrate both sides.

$\int \mathsf{e^{x\:sinx+cosx}\big[\frac{x^{4}\:cos^{3}x-x\:sinx+cosx}{x^{2}\:cos^{2}x}\big]\:dx=\int dt}$

Hence, the required integral is:

$\mathsf{\big(x-\frac{1}{x\:cosx}\big).e^{x\:sinx+cosx}+C}$