If I=∫exsin⁡x+cos⁡xx4cos3⁡x−xsin⁡x+cos⁡xx2cos2⁡xdx then I equal to.

I=∫exsin⁡x+cos⁡xx2cos⁡x+cos⁡x−xsin⁡x(xcos⁡x)2dx=I1+I2where I1=∫exsin⁡x+cos⁡xxcos⁡xxdxNoting ddx(xsin⁡x+cos⁡x)=xcos⁡x, we get I1=xexsin⁡x+cos⁡x−∫exsin⁡x+cos⁡xdx and I2=∫exsin⁡x+cos⁡x(cos⁡x−xsin⁡x)(xcos⁡x)2dx=exsin⁡x+cos⁡x−1(xcos⁡x)+∫exsin⁡x+cos⁡x(xcos⁡x)dxxcos⁡xThus I=exsin⁡x+cos⁡xx−1xcos⁡x+c

If I=∫exsin⁡x+cos⁡xx4cos3⁡x−xsin⁡x+cos⁡xx2cos2⁡xdx then I equal to.

I=∫exsin⁡x+cos⁡xx2cos⁡x+cos⁡x−xsin⁡x(xcos⁡x)2dx=I1+I2where I1=∫exsin⁡x+cos⁡xxcos⁡xxdxNoting ddx(xsin⁡x+cos⁡x)=xcos⁡x, we get I1=xexsin⁡x+cos⁡x−∫exsin⁡x+cos⁡xdx and I2=∫exsin⁡x+cos⁡x(cos⁡x−xsin⁡x)(xcos⁡x)2dx=exsin⁡x+cos⁡x−1(xcos⁡x)+∫exsin⁡x+cos⁡x(xcos⁡x)dxxcos⁡xThus I=exsin⁡x+cos⁡xx−1xcos⁡x+c

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If $I = \int e^{x \sin x + \cos x} (\frac{x^{4} \cos^{3} x - x \sin x + \cos x}{x^{2} \cos^{2} x}) d x$ then $I$ equal to.

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$e^{x \sin x + \cos x} (x - \frac{\sec x}{x}) + C$

$e^{x \sin x + \cos x} (x \sin x - \frac{\cos x}{x})$

$e^{x \sin x + \cos x} (\frac{x}{\tan x} - \frac{\sec x}{x}) + C$

$x e^{x \sin x + \cos x} - \int e^{x \sin x + \cos x} (1 - \frac{\cos x - x \sin x}{x^{2} \cos^{2} x}) d x$

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int e^{x \sin x + \cos x} \{x^{2} \cos x + \frac{\cos x - x \sin x}{(x \cos x)^{2}}\} d x \\ = I_{1} + I_{2} \end{array}$

where $I_{1} = \int (e^{x \sin x + \cos x} x \cos x) x d x$

Noting $\frac{d}{d x} (x \sin x + \cos x) = x \cos x, we get$

$\begin{aligned} I_{1} = x e^{x \sin x + \cos x} - \int e^{x \sin x + \cos x} d x and \\ I_{2} = \int e^{x \sin x + \cos x} \frac{(\cos x - x \sin x)}{(x \cos x)^{2}} d x \end{aligned}$