If f(x) = int x^2 cos^2 x (2x tan^2 x - 2x - | vedclass

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(C) Given $f(x) = \int x^2 \cos^2 x (2x 	an^2 x - 2x - 6 	an x) dx$. Expanding the integrand: $f(x) = \int (2x^3 \sin^2 x - 2x^3 \cos^2 x - 6x^2 \si

Vedclass · Accepted Answer

$-x^3 \sin 2x + \pi$

Vedclass · Answer

(C) Given $f(x) = \int x^2 \cos^2 x (2x 	an^2 x - 2x - 6 	an x) dx$. Expanding the integrand: $f(x) = \int (2x^3 \sin^2 x - 2x^3 \cos^2 x - 6x^2 \sin x \cos x) dx$. Using trigonometric identities $\sin^2 x - \cos^2 x = -\cos 2x$ and $2 \sin x \cos x = \sin 2x$: $f(x) = \int (-2x^3 \cos 2x - 3x^2 \sin 2x) dx$. Let $I = \int (-2x^3 \cos 2x - 3x^2 \sin 2x) dx$. Using integration by parts on the first term $\int u dv = uv - \int v du$ where $u = -x^3$ and $dv = 2 \cos 2x dx$: $I = -x^3 \sin 2x - \int (-\sin 2x)(3x^2) dx - \int 3x^2 \sin 2x dx$. $I = -x^3 \sin 2x + \int 3x^2 \sin 2x dx - \int 3x^2 \sin 2x dx$. $I = -x^3 \sin 2x + C$. Given $f(0) = \pi$,we have $-(0)^3 \sin(0) + C = \pi$,which implies $C = \pi$. Therefore,$f(x) = -x^3 \sin 2x + \pi$.

List-$I$	List-$II$
$1. \int \frac{\sin^2 x}{\cos^4 x} dx$	$A. \frac{\tan^2 x}{2} + \ln\|\cos x\| + c$
$2. \int \frac{\sin^4 x}{\cos^2 x} dx$	$B. \cos x + \sec x + c$
$3. \int \frac{\sin^3 x}{\cos^2 x} dx$	$C. \frac{\tan^3 x}{3} + c$
$4. \int \frac{\sin^3 x}{\cos^3 x} dx$	$D. \tan x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$
	$E. \cos x - \sec x + c$

If $f(x) = \int x^2 \cos^2 x (2x \tan^2 x - 2x - 6 \tan x) dx$ and $f(0) = \pi$,then $f(x) =$

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