≤ft(1/x²+³ x+³ x/² x² x)\,dx =

Correct Answer is : (sin⁡x−cos⁡x)x−sin⁡xcos⁡xxsin⁡xcos⁡x+C( x- x)x- x x/x x x+Cxsinxcosx(sinx−cosx)x−sinxcosx+C

Correct Answer is : (sin⁡x−cos⁡x)x−sin⁡xcos⁡xxsin⁡xcos⁡x+C( x- x)x- x x/x x x+Cxsinxcosx(sinx−cosx)x−sinxcosx+C

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TG EAMCET 2 May 2025 Shift 2 Paper

Question : 72 of 160

Marks: +1, -0

\int\left(\frac{1}{x^2}+\frac{\sin^3 x+\cos^3 x}{\sin^2 x\cos^2 x}\right)\,dx =

Solution:

\text{ Let } I=\int\left(\frac{1}{x^2}+\frac{\sin^3 x+\cos^3 x}{\sin^2 x\cos^2 x}\right)\,dx

\text{ again let } I_1=\int\frac{1}{x^2}\,dx

\text{ And } I_2=\int\frac{\sin^3 x+\cos^3 x}{\sin^2 x\cos^2 x}\,dx

\therefore I_1=\int x^{-2}\,dx=\frac{x^{-2+1}}{-2+1}+c_1=-\frac{1}{x}+c_1

\text{ And } I_2=\int\left(\frac{\sin^3 x}{\sin^2 x\cos^2 x}+\frac{\cos^3 x}{\sin^2 x\cos^2 x}\right)\,dx

=\int(\tan x \sec x+\cot x \csc x)\,dx

=\int\tan x \sec x\,dx+\int\cot x \csc x\,dx

=\sec x+(-\csc x)+c_2

\therefore I=I_1+I_2=-\frac{1}{x}+\sec x-\csc x+C

\Rightarrow I=-\frac{1}{x}+\frac{1}{\cos x}-\frac{1}{\sin x}+C

=\frac{-(\cos x \sin x)+x \sin x-x \cos x}{x \cos x \sin x}+C

=\frac{x(\sin x-\cos x)-\cos x \sin x}{x \cos x \sin x}+C

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