int frac{sin^3 2x}{cos^5 2x} , dx = | vedclass

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Question

(D) Let $I = \int \frac{\sin^3 2x}{\cos^5 2x} \, dx$. We can rewrite the integrand as: $I = \int \frac{\sin^3 2x}{\cos^3 2x} \cdot \frac{1}{\cos^2 2x}

Vedclass · Accepted Answer

$\frac{1}{8} 	an^4 2x + C$

Vedclass · Answer

(D) Let $I = \int \frac{\sin^3 2x}{\cos^5 2x} \, dx$. We can rewrite the integrand as: $I = \int \frac{\sin^3 2x}{\cos^3 2x} \cdot \frac{1}{\cos^2 2x} \, dx = \int 	an^3 2x \cdot \sec^2 2x \, dx$. Now,use the substitution method. Let $u = 	an 2x$. Then,the derivative is $du = 2 \sec^2 2x \, dx$,which implies $\sec^2 2x \, dx = \frac{du}{2}$. Substituting these into the integral: $I = \int u^3 \cdot \frac{du}{2} = \frac{1}{2} \int u^3 \, du$. Integrating $u^3$ with respect to $u$ gives $\frac{u^4}{4}$. Thus,$I = \frac{1}{2} \cdot \frac{u^4}{4} + C = \frac{1}{8} u^4 + C$. Substituting back $u = 	an 2x$,we get: $I = \frac{1}{8} 	an^4 2x + C$.

$\int \frac{\sin^3 2x}{\cos^5 2x} \, dx = $

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