If tan alpha = frac{4}{3},then int frac{1}{3 | vedclass

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(B) Given $	an \alpha = \frac{4}{3}$. Since $	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{4}{3}$,we have $\sin \alpha = \frac{4}{5}$ and $\c

Vedclass · Accepted Answer

$\frac{1}{5} \log \left| 	an \left( \frac{\pi}{4} + \frac{x}{2} + \frac{\alpha}{2} ight) ight| + c$

Vedclass · Answer

(B) Given $	an \alpha = \frac{4}{3}$. Since $	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{4}{3}$,we have $\sin \alpha = \frac{4}{5}$ and $\cos \alpha = \frac{3}{5}$.Let $I = \int \frac{dx}{3 \cos x - 4 \sin x}$.Multiply and divide by $\sqrt{3^2 + 4^2} = 5$:$I = \frac{1}{5} \int \frac{dx}{\frac{3}{5} \cos x - \frac{4}{5} \sin x} = \frac{1}{5} \int \frac{dx}{\cos \alpha \cos x - \sin \alpha \sin x}$.Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:$I = \frac{1}{5} \int \frac{dx}{\cos(x + \alpha)} = \frac{1}{5} \int \sec(x + \alpha) dx$.The integral of $\sec 	heta$ is $\log | \sec 	heta + 	an 	heta |$ or $\log | 	an(\frac{\pi}{4} + \frac{	heta}{2}) |$.Thus,$I = \frac{1}{5} \log \left| 	an \left( \frac{\pi}{4} + \frac{x + \alpha}{2} ight) ight| + c = \frac{1}{5} \log \left| 	an \left( \frac{\pi}{4} + \frac{x}{2} + \frac{\alpha}{2} ight) ight| + c$.

If $\tan \alpha = \frac{4}{3}$,then $\int \frac{1}{3 \cos x - 4 \sin x} dx = $

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