The value of int frac{sin^3 x}{(cos^4 x + 3 c | vedclass

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Question

(A) Let $I = \int \frac{\sin^3 x}{(\cos^4 x + 3 \cos^2 x + 1) 	an^{-1}(\sec x + \cos x)} dx$.Substitute $u = \sec x + \cos x$. Then $du = (\sec x 	a

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$\log(	an^{-1}(\sec x + \cos x)) + c$

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(A) Let $I = \int \frac{\sin^3 x}{(\cos^4 x + 3 \cos^2 x + 1) 	an^{-1}(\sec x + \cos x)} dx$.Substitute $u = \sec x + \cos x$. Then $du = (\sec x 	an x - \sin x) dx = (\frac{\sin x}{\cos^2 x} - \sin x) dx = \sin x (\frac{1 - \cos^2 x}{\cos^2 x}) dx = \sin x 	an^2 x dx$.Note that $\sin^3 x dx = \sin x (1 - \cos^2 x) dx$. This does not simplify directly. Let us rewrite the denominator term: $\cos^4 x + 3 \cos^2 x + 1 = \cos^2 x (\cos^2 x + 3 + \sec^2 x) = \cos^2 x ((\sec x + \cos x)^2 + 1)$.Thus, $I = \int \frac{\sin x (1 - \cos^2 x)}{\cos^2 x (u^2 + 1) 	an^{-1} u} dx = \int \frac{\sin x (\sec^2 x - 1)}{(u^2 + 1) 	an^{-1} u} dx$.Since $du = \sin x 	an^2 x dx = \sin x (\sec^2 x - 1) dx$, we have $I = \int \frac{du}{(u^2 + 1) 	an^{-1} u}$.Let $v = 	an^{-1} u$, then $dv = \frac{1}{u^2 + 1} du$.Therefore, $I = \int \frac{1}{v} dv = \log|v| + c = \log(	an^{-1}(\sec x + \cos x)) + c$.

The value of $\int \frac{\sin^3 x}{(\cos^4 x + 3 \cos^2 x + 1) \tan^{-1}(\sec x + \cos x)} dx$ is

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