If int frac{3 sin x cos x}{4 sin x+7} , dx = | vedclass

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(C) Let $I = \int \frac{3 \sin x \cos x}{4 \sin x + 7} \, dx$. Substitute $u = \sin x$,then $du = \cos x \, dx$. The integral becomes $I = \int \frac{

Vedclass · Accepted Answer

$\frac{33}{16}$

Vedclass · Answer

(C) Let $I = \int \frac{3 \sin x \cos x}{4 \sin x + 7} \, dx$. Substitute $u = \sin x$,then $du = \cos x \, dx$. The integral becomes $I = \int \frac{3u}{4u+7} \, du$. Perform polynomial division or rewrite the numerator: $\frac{3u}{4u+7} = \frac{3}{4} \left( \frac{4u}{4u+7} \right) = \frac{3}{4} \left( \frac{4u+7-7}{4u+7} \right) = \frac{3}{4} \left( 1 - \frac{7}{4u+7} \right) = \frac{3}{4} - \frac{21}{4(4u+7)}$. Integrating with respect to $u$: $I = \int \left( \frac{3}{4} - \frac{21}{4(4u+7)} \right) \, du = \frac{3}{4}u - \frac{21}{4} \cdot \frac{1}{4} \log |4u+7| + c$. Substituting back $u = \sin x$: $I = \frac{3}{4} \sin x - \frac{21}{16} \log |4 \sin x + 7| + c$. Comparing this with $A \sin x - B \log |4 \sin x + 7| + c$,we get $A = \frac{3}{4}$ and $B = \frac{21}{16}$. Thus,$A+B = \frac{3}{4} + \frac{21}{16} = \frac{12+21}{16} = \frac{33}{16}$.

If $\int \frac{3 \sin x \cos x}{4 \sin x+7} \, dx = A \sin x - B \log |4 \sin x + 7| + c$ where $c$ is the constant of integration,then the value of $A+B$ is equal to

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